Re: [Tutor] Dictionary viceversa

[Alan Gauld via Tutor]

On 30/07/18 19:11, Zachary Ware wrote:
> On Mon, Jul 30, 2018 at 1:08 PM Alan Gauld via Tutor  wrote:
>> There are lots of options including those suggested elsewhere.
>> Another involves using get() which makes your function
>> look like:
>>
>> def viceversa(d):
>> new_d = dict()
>> for k in d:
>> for e in d[k]:
>> new_d[e] = new_d.get(e,[]).append(k)
> 
> Note that this will set each entry to `None` as returned by `list.append`.


Oops, yes. You need an intermediate variable:

 for e in d[k]:
 data = new_d.get(e,[])
 data.append(k)
 new_d[e] = data

Or use addition:

 for e in d[k]:
 new_d[e] = new_d.get(e,[]) + [k]


-- 
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary viceversa

[Zachary Ware]

On Mon, Jul 30, 2018 at 1:08 PM Alan Gauld via Tutor  wrote:
> There are lots of options including those suggested elsewhere.
> Another involves using get() which makes your function
> look like:
>
> def viceversa(d):
> new_d = dict()
> for k in d:
> for e in d[k]:
> new_d[e] = new_d.get(e,[]).append(k)

Note that this will set each entry to `None` as returned by `list.append`.

-- 
Zach
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Re: [Tutor] Dictionary viceversa

[Alan Gauld via Tutor]

On 30/07/18 13:40, Valerio Pachera wrote:

> users = {'user1':['office-a', 'office-b'],
>  'user2':['office-b'],
>  'user3':['office-a','office-c']}
> 
> It's a list of users.
> For each user there's a list of room it can access to.
> 
> I wish to get the same info but "sorted" by room.

Remember that dicts are not sorted. You can create a dictionary
keyed by room but not sorted by room.
(You can however get a list of sorted keys but that's different,
and you can use a collections.OrderedDict)

> And i generalized it in a function like this:
> 
> def viceversa(d):
> new_d = dict()
> for k in d:
> for e in d[k]:
> if e in new_d:
> new_d[e].append(k)
> else:
> new_d[e] = []
> new_d[e].append(k)
> return(new_d)
> 
> My question is: is there a better way to that?

There are lots of options including those suggested elsewhere.
Another involves using get() which makes your function
look like:

def viceversa(d):
new_d = dict()
for k in d:
for e in d[k]:
new_d[e] = new_d.get(e,[]).append(k)
return(new_d)

> Maybe by list comprehension?

I can't think of a reasonable way of doing that but a
generator may work.

Another option would be to build a set of the original
rooms then iterate over the data, collecting the keys
which have those rooms. But that would be pretty
inefficient...

-- 
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary viceversa

[Peter Otten]

Zachary Ware wrote:

> On Mon, Jul 30, 2018 at 12:20 PM Valerio Pachera  wrote:
>> I was looking to substiture the cicle for e in new_d like this:
>>   [ new_d[e].append(k) if e in new_d else new_d[e].append(k) for e in
>>   [ d[k] ]
>> but it can't work because 'new_d[e] = []' is missing.
> 
> Have a look at `dict.setdefault` and play around with it; I think it
> will help you do what you want.

Either that, or use a defaultdict:
 
>>> import collections
>>> room_access = collections.defaultdict(list)
>>> for user, rooms in users.items():
... for room in rooms:
... room_access[room].append(user)
... 
>>> room_access
defaultdict(, {'office-c': ['user3'], 'office-b': ['user2', 
'user1'], 'office-a': ['user3', 'user1']})


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Re: [Tutor] Dictionary viceversa

[Zachary Ware]

On Mon, Jul 30, 2018 at 12:20 PM Valerio Pachera  wrote:
> I was looking to substiture the cicle for e in new_d like this:
>   [ new_d[e].append(k) if e in new_d else new_d[e].append(k) for e in d[k] ]
> but it can't work because 'new_d[e] = []' is missing.

Have a look at `dict.setdefault` and play around with it; I think it
will help you do what you want.

-- 
Zach
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Re: [Tutor] Dictionary Question

[Michael Selik]

On Mon, May 2, 2016 at 5:28 PM Jason N. via Tutor  wrote:

> Hello,
> Wanted to ask if its possible to have a dictionary that can be looked up
> by either values?
> For example,
> mydic = {"A: "Apple", "B": "Banana"}When user inputs "A" I want "Apple" to
> come. But if the user enter "Apple" I want "A" to respond.
> Please let me know the best way to handle this type cause instead of just
> created duplicate entries to cover all possibilities. Thank you.
>

A dictionary enforces that the keys are unique, but many keys may have the
same value. Do you want to enforce that values are unique? If not, does it
matter which key is returned if many keys have the same value?
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Re: [Tutor] Dictionary Question

[Michael Selik]

On Mon, May 2, 2016 at 8:58 PM Jason N. via Tutor  wrote:

> What is the best way to make dictionary requests case in-sensitive? For
> example, "Apple and "apple" should bring back the same dictionary
> response. Thank you.
>

Take a look at how the requests library solves the problem with a
"CaseInsensitiveDict" (
https://github.com/kennethreitz/requests/blob/master/requests/structures.py)
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Re: [Tutor] Dictionary Question

[bharath ks via Tutor]

Hello,

Using iteritems would be much easier approach
Something like this
mydic = {"A": "Apple", "B": "Banana"}
for key, value in mydic.iteritems():    if value == "Apple":        print key








 Thanks & BR, Bharath Shetty 

On Tuesday, 3 May 2016 2:57 AM, Jason N. via Tutor  wrote:
 

 Thank you all for your responses. 
A quick follow up, what is the best way to make dictionary requests case 
in-sensitive? For example, "Apple and "apple" should bring back the same 
dictionary response. Thank you. 

    On Monday, May 2, 2016 6:57 PM, Bob Gailer  wrote:
 
 

 
On May 2, 2016 5:27 PM, "Jason N. via Tutor"  wrote:
>
> Hello,
> Wanted to ask if its possible to have a dictionary that can be looked up by 
> either values?
> For example, 
> mydic = {"A: "Apple", "B": "Banana"}When user inputs "A" I want "Apple" to 
> come. But if the user enter "Apple" I want "A" to respond.
I think this would depend on how big the data set is and how often you want to 
look things up.
Two other Solutions: 
Create a class which internally manages two dictionaries.
If things are really big create a database using for example sqlite.

 
  
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Re: [Tutor] Dictionary Question

[cs]

On 03May2016 00:56, Jason N.  wrote:

Thank you all for your responses. 
A quick follow up, what is the best way to make dictionary requests case 
in-sensitive? For example, "Apple and "apple" should bring back the same 
dictionary response. Thank you.


There are a few ways depending what your more fine grained objectives are. But 
they all tend to revolve around "normalising" the keys, which is a common 
practice for many things where multiple values are considered the same: in your 
case upper and lower case.


So the easy thing is to always convert to lower case (or upper case, but lower 
case is less SHOUTY). Eg:


 def save(d, key, value):
   d[key.lower()] = value

so the normalising function here is d.lower.

Usually you'd be making yourself a mapping class of some kind: an object which 
behaves like a dictionay:


 https://docs.python.org/3/glossary.html#term-mapping

And internally it would usually have a dictionary for storage. Completely 
untested example code:


 class CaseInsensitiveMapping:

   def __init__(self):
 self._d = {}

   def __getitem__(self, key):
 return self._d[key.lower()]
   
   def __setitem__(self, key, value):

 self._d[key.lower()] = value

and so forth for the other special ("dunder" in Pythonspeak) methods used to 
implement a mapping:


 https://docs.python.org/3/reference/datamodel.html#emulating-container-types

From the outside:

 cimap = CaseInsensitiveMapping()
 cimap['X']=1
 print(cimap['x'])

should print 1.

Now having sketched a trivial example like this, you might need to be more 
elaborate depending on youruse case. For example, some mappings like this one 
preserve the case used to insert the original key. So while ['X'] and ['x'] 
would both find the value 1, they .keys() method with recite 'X' because that 
was the specific string used to put the 1 into the mapping. That would make the 
internal implementation more complicated.


Cheers,
Cameron Simpson 
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Re: [Tutor] Dictionary Question

[isaac tetteh]

If only I understand what you mean. You can just make all the values in the 
dictionary lower, upper or capitalized. Then if you want take an input or 
whatever you want to do with it just use .lower() or .upper() or .capitalized() 
to convert it to what is in the dictionary. Maybe someone has a better way to 
do it :) 


Sent from my iPhone

> On May 2, 2016, at 7:59 PM, Jason N. via Tutor  wrote:
> 
> Thank you all for your responses. 
> A quick follow up, what is the best way to make dictionary requests case 
> in-sensitive? For example, "Apple and "apple" should bring back the same 
> dictionary response. Thank you. 
> 
>On Monday, May 2, 2016 6:57 PM, Bob Gailer  wrote:
> 
> 
> 
> 
>> On May 2, 2016 5:27 PM, "Jason N. via Tutor"  wrote:
>> 
>> Hello,
>> Wanted to ask if its possible to have a dictionary that can be looked up by 
>> either values?
>> For example, mydic = {"A: "Apple", "B": "Banana"}
>> When user inputs "A" I want "Apple" to come. But if the user enter "Apple" I 
>> want "A" to respond.
> I think this would depend on how big the data set is and how often you want 
> to look things up.
> Two other Solutions: 
> Create a class which internally manages two dictionaries.
> If things are really big create a database using for example sqlite.
> 
> 
> 
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Re: [Tutor] Dictionary Question

[Jason N. via Tutor]

Thank you all for your responses. 
A quick follow up, what is the best way to make dictionary requests case 
in-sensitive? For example, "Apple and "apple" should bring back the same 
dictionary response. Thank you. 

On Monday, May 2, 2016 6:57 PM, Bob Gailer  wrote:
 
 

 
On May 2, 2016 5:27 PM, "Jason N. via Tutor"  wrote:
>
> Hello,
> Wanted to ask if its possible to have a dictionary that can be looked up by 
> either values?
> For example, 
> mydic = {"A: "Apple", "B": "Banana"}When user inputs "A" I want "Apple" to 
> come. But if the user enter "Apple" I want "A" to respond.
I think this would depend on how big the data set is and how often you want to 
look things up.
Two other Solutions: 
Create a class which internally manages two dictionaries.
If things are really big create a database using for example sqlite.

 
  
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Re: [Tutor] Dictionary Question

[Bob Gailer]

On May 2, 2016 5:27 PM, "Jason N. via Tutor"  wrote:
>
> Hello,
> Wanted to ask if its possible to have a dictionary that can be looked up
by either values?
> For example,
> mydic = {"A: "Apple", "B": "Banana"}When user inputs "A" I want "Apple"
to come. But if the user enter "Apple" I want "A" to respond.
I think this would depend on how big the data set is and how often you want
to look things up.
Two other Solutions:
Create a class which internally manages two dictionaries.
If things are really big create a database using for example sqlite.
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Re: [Tutor] Dictionary Question

[Alan Gauld via Tutor]

On 02/05/16 22:55, isaac tetteh wrote:
> 
> For some reason i cant find reply all . But try this 
> for key, value in mydic.items():
>   If A==value:
>Print key

or as a function:

def findKey(dct, val):
   for k,v in dct.items():
  if v == val:
 return k

mydic = {"A: "Apple", "B": "Banana"}

print( findKey(mydic,'Apple') )   # -> 'A'

The problem is that while keys are unique, values
might not be, so what do you do if multiple keys
share the same value?

You could use a comprehension:

def findKeys(dct,val):
keys = [k for k,v in dct.items() if v == val]
return keys

But if you are only interested in one of them then
it's down to you to figure out which!

-- 
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Question

[isaac tetteh]

Sorry for the if statement the correct statement should be "if 'apple' ==value:"

Sent from my iPhone

> On May 2, 2016, at 4:58 PM, isaac tetteh  wrote:
> 
> 
> For some reason i cant find reply all . But try this 
> for key, value in mydic.items():
>  If A==value:
>   Print key
> Nb: use iteritems() if using python2
> 
> Sent from my iPhone
> 
>> On May 2, 2016, at 4:29 PM, Jason N. via Tutor  wrote:
>> 
>> Hello,
>> Wanted to ask if its possible to have a dictionary that can be looked up by 
>> either values?
>> For example, 
>> mydic = {"A: "Apple", "B": "Banana"}When user inputs "A" I want "Apple" to 
>> come. But if the user enter "Apple" I want "A" to respond.
>> Please let me know the best way to handle this type cause instead of just 
>> created duplicate entries to cover all possibilities. Thank you.
>> ___
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Re: [Tutor] Dictionary Question

[isaac tetteh]

For some reason i cant find reply all . But try this 
for key, value in mydic.items():
  If A==value:
   Print key
Nb: use iteritems() if using python2

Sent from my iPhone

> On May 2, 2016, at 4:29 PM, Jason N. via Tutor  wrote:
> 
> Hello,
> Wanted to ask if its possible to have a dictionary that can be looked up by 
> either values?
> For example, 
> mydic = {"A: "Apple", "B": "Banana"}When user inputs "A" I want "Apple" to 
> come. But if the user enter "Apple" I want "A" to respond.
> Please let me know the best way to handle this type cause instead of just 
> created duplicate entries to cover all possibilities. Thank you.
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Re: [Tutor] Dictionary on data

[Peter Otten]

jarod_v6--- via Tutor wrote:

> Dear All!
> I have this  elements
> 
> In [445]: pt = line.split("\t")[9]
> 
> In [446]: pt
> Out[446]: 'gene_id "ENSG0223972"; gene_version "5"; transcript_id
> "ENST0456328"; transcript_version "2"; exon_number "1"; gene_name
> "DDX11L1"; gene_source "havana"; gene_biotype
> "transcribed_unprocessed_pseudogene"; transcript_name "DDX11L1-002";
> transcript_source "havana"; transcript_biotype "processed_transcript";
> exon_id "ENSE2234944"; exon_version "1"; tag "basic";
> transcript_support_level "1";\n'
> 
> 
> and I want to create a dictionary like this
> 
> gene_id =  "ENSG0223972"; ...
> 
> 
> I found on stack over flow this way to create a dictionary of dictionary
> (http://stackoverflow.com/questions/8550912/python-dictionary-of-dictionaries)
> # This is our sample data
> data = [("Milter", "Miller", 4), ("Milter", "Miler", 4), ("Milter",
> "Malter", 2)]
> 
> # dictionary we want for the result
> dictionary = {}
> 
> # loop that makes it work
>  for realName, falseName, position in data:
> dictionary.setdefault(realName, {})[falseName] = position
> 
> I want to create a dictionary using   setdefault but I have difficult to
> trasform pt as list of tuple.
> 
>  data = pt.split(";")
>  in ()
>   1 for i in data:
>   2 l = i.split()
> > 3 print l[0]
>   4
> 
> IndexError: list index out of range
> 
> In [457]: for i in data:
> l = i.split()
> print l
>.:
> ['gene_id', '"ENSG0223972"']
> ['gene_version', '"5"']
> ['transcript_id', '"ENST0456328"']
> ['transcript_version', '"2"']
> ['exon_number', '"1"']
> ['gene_name', '"DDX11L1"']
> ['gene_source', '"havana"']
> ['gene_biotype', '"transcribed_unprocessed_pseudogene"']
> ['transcript_name', '"DDX11L1-002"']
> ['transcript_source', '"havana"']
> ['transcript_biotype', '"processed_transcript"']
> ['exon_id', '"ENSE2234944"']
> ['exon_version', '"1"']
> ['tag', '"basic"']
> ['transcript_support_level', '"1"']
> []
> 
> 
> So how can do that more elegant way?
> thanks so much!!

I don't see why you would need dict.setdefault(), you have the necessary 
pieces together:

data = pt.split(";")
pairs = (item.split() for item in data)
mydict = {item[0]: item[1].strip('"') for item in pairs if len(item) == 2}

You can protect against whitespace in the quoted strings with 
item.split(None, 1) instead of item.split(). If ";" is allowed in the quoted 
strings you have to work a little harder.



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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 08/08/15 00:05, Ltc Hotspot wrote:

Hi Alan,

On line 15, I replaced: 'count[address] = count.get(address, 0) + 1' 
with 'line = Counter(address)'.


line = Counter(address)

will create a new Counter object with the address in it with a count 
value of 1.
Every line in the file will create a new Counter, overwriting the 
previous one.
At the end of the loop you will have exactly 1 Counter containing the 
last item

in the loop and a count of 1.

Instead create the Counter before the loop and then use the update()
method to add the address to it. Recall we talked about reading the Counter
documentation? It describes the update() method.

One you have the Counter with all the addresses in you can get the max 
value

and key directly without using a second loop and your maxkee and maxval
variables.

Again read the Counter documentation to see how.

Finally remember the imports. You never show us those but the Counter will
not work unless you import it.


Question: What is the cause of the empty dict?


You create it but never use it. You replaced the line where count got
populated with your Counter call. So count never gets populated.
Just because Counter sounds like count does not mean they are
in any way connected.

--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos

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Re: [Tutor] Dictionary Issue

[Emile van Sebille]

On 8/6/2015 5:30 PM, Ltc Hotspot wrote:


I'm following the instructor's video exercise, available  at
https://www.youtube.com/watch?v=3cwXN5_3K6Q.


As you're clearly interested in learning python, you may find working 
the tutorial beneficial as it steps you through the fundamentals of 
python in a time tested way.


See https://docs.python.org/2/tutorial/ for python2 or
https://docs.python.org/3/tutorial/ for python3.

Emile



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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

Alan,

I want to find the max val ,  keys and values are defined on line 10:
for kee, val in count.items():

Then,  maxval determines the max value on line 11:
if val  maxval:

right, view a copy of the revised code at
http://tinyurl.com/nvzdw8k

Question1: are these assumptions true, above?

Question2: dict maps strings into keys and values?


Hal



On Thu, Aug 6, 2015 at 11:35 PM, Alan Gauld alan.ga...@btinternet.com
wrote:

 On 07/08/15 01:15, Ltc Hotspot wrote:

 Question1: How type of argument should I use for dict, i.e.,user argument
 or list argument.

 Read captured traceback:

 TypeError
 Traceback (most recent call last)
 C:\Users\vm\Desktop\apps\docs\Python\assignment_9_4_26.py in module()
1 fname = raw_input(Enter file name: )
2 handle = open (fname, 'r')
  3 count = dict.keys()
4 for line in handle:
5 if line.startswith(From: ):



 You appear to be making random changes to your code
 for no good reason.

 I will not make any further suggestions until you
 start to explain your thinking.

 What do you think the line

 count = dict.keys()

 will do? Why do you want to do that?
 How will it help you solve your problem?

 --
 Alan G
 Author of the Learn to Program web site
 http://www.alan-g.me.uk/
 http://www.amazon.com/author/alan_gauld
 Follow my photo-blog on Flickr at:
 http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

Mark,

I'm following the instructor's video exercise, available  at
https://www.youtube.com/watch?v=3cwXN5_3K6Q.

View attached screen shot file, image file shows a copy of the
counter: cou[wrd] =cou.get(wrd,0) +1


Please, explain the differences in counter methods?


Hal

On Thu, Aug 6, 2015 at 4:53 PM, Mark Lawrence breamore...@yahoo.co.uk
wrote:

 On 06/08/2015 20:05, Ltc Hotspot wrote:

 On my breath and soul, I did:

 Counter objects have a dictionary interface except that they return a zero
 count for missing items instead of raising a KeyError
 https://docs.python.org/3/library/exceptions.html#KeyError:


 That's nice to know.  What do the rest of the methods on the class do?

 Please don't top post here, it makes following long threads difficult.


 What did you not understand about the above?

 You obviously haven't bothered to read the link I gave you about the
 Counter class so I give up.


 If you'd read the entire write up why are you still wasting time with a
 loop to find a maximum that simply doesn't work, when there is likely a
 solution in the Counter class right in front of your eyes?


 --
 My fellow Pythonistas, ask not what our language can do for you, ask
 what you can do for our language.

 Mark Lawrence

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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

Question1: How type of argument should I use for dict, i.e.,user argument
or list argument.

Read captured traceback:

TypeError
Traceback (most recent call last)
C:\Users\vm\Desktop\apps\docs\Python\assignment_9_4_26.py in module()
  1 fname = raw_input(Enter file name: )
  2 handle = open (fname, 'r')
 3 count = dict.keys()
  4 for line in handle:
  5 if line.startswith(From: ):

TypeError: descriptor 'keys' of 'dict' object needs an argument

In [99]:

Question2: Are all the loop failures resolved in the revised code?

Revised code is available at
https://gist.github.com/ltc-hotspot/00fa77ca9b40c0a77170

Regards,
Hal

On Thu, Aug 6, 2015 at 4:20 PM, Alan Gauld alan.ga...@btinternet.com
wrote:

 On 07/08/15 00:11, Ltc Hotspot wrote:

 Questions(1):Why does print line, prints blank space; and, (2) print
 address prints a single email address:


 See my previous emails.
 You are not storing your addresses so address only holds the last address
 in the file.
 line is at the end of the file so is empty.,

 In [72]: print count
 {'gopal.ramasammyc...@gmail.com mailto:gopal.ramasammyc...@gmail.com':
 1, 'lo...@media.berkeley.edu mailto:lo...@media.berkeley.edu': 3,
 'cwen@iupui.
 edu': 5, 'antra...@caret.cam.ac.uk mailto:antra...@caret.cam.ac.uk':
 1, 'rjl...@iupui.edu mailto:rjl...@iupui.edu': 2, 'gsil...@umich.ed
 u': 3, 'david.horw...@uct.ac.za mailto:david.horw...@uct.ac.za': 4, '
 wagne...@iupui.edu mailto:wagne...@iupui.edu': 1, 'zq...@umich.edu
 mailto:zq...@umich.edu':
  4, 'stephen.marqu...@uct.ac.za mailto:stephen.marqu...@uct.ac.za': 2,
 'r...@media.berkeley.edu mailto:r...@media.berkeley.edu': 1}

 Question(3): why did the elements print count('keys') and print
 count('items') fail?


 Because, as shown above, count is a dictionary.
 So items and keys are methods not strings to be passed
 to a non-existent count() function.

 So you need, for example:

 print count.keys()

 Traceback (most recent call last)
 ipython-input-76-35c8707b256e in module()
  1 print count('items')

 TypeError: 'dict' object is not callable


 Which is what the error is also telling you.
 You cannot call - ie use () - with a dictionary like count.


 --
 Alan G
 Author of the Learn to Program web site
 http://www.alan-g.me.uk/
 http://www.amazon.com/author/alan_gauld
 Follow my photo-blog on Flickr at:
 http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 07/08/15 01:15, Ltc Hotspot wrote:

Question1: How type of argument should I use for dict, i.e.,user argument
or list argument.

Read captured traceback:

TypeError
Traceback (most recent call last)
C:\Users\vm\Desktop\apps\docs\Python\assignment_9_4_26.py in module()
   1 fname = raw_input(Enter file name: )
   2 handle = open (fname, 'r')
 3 count = dict.keys()
   4 for line in handle:
   5 if line.startswith(From: ):



You appear to be making random changes to your code
for no good reason.

I will not make any further suggestions until you
start to explain your thinking.

What do you think the line

count = dict.keys()

will do? Why do you want to do that?
How will it help you solve your problem?

--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Mark Lawrence]

On 07/08/2015 01:30, Ltc Hotspot wrote:

Mark,

I'm following the instructor's video exercise, available  at
https://www.youtube.com/watch?v=3cwXN5_3K6Q.

View attached screen shot file, image file shows a copy of the
counter: cou[wrd] =cou.get(wrd,0) +1

Please, explain the differences in counter methods?



top posted, again, *plonk*

--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence

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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

Hi Mark,

Why is Counter not defined on line #15: line =
Counter(address),i.e., NameError: name 'Counter' is not defined?

Share a chat session at http://tinyurl.com/oull2fw
View line entry at http://tinyurl.com/oggzn97

Hal

On Fri, Aug 7, 2015 at 12:14 AM, Mark Lawrence breamore...@yahoo.co.uk
wrote:

 On 07/08/2015 01:30, Ltc Hotspot wrote:

 Mark,

 I'm following the instructor's video exercise, available  at
 https://www.youtube.com/watch?v=3cwXN5_3K6Q.

 View attached screen shot file, image file shows a copy of the
 counter: cou[wrd] =cou.get(wrd,0) +1

 Please, explain the differences in counter methods?


 top posted, again, *plonk*


 --
 My fellow Pythonistas, ask not what our language can do for you, ask
 what you can do for our language.

 Mark Lawrence

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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 07/08/15 19:18, Ltc Hotspot wrote:


I want to find the max val ,  keys and values are defined on line 10:
for kee, val in count.items():


Yes we know that, but you are not answering the questions we pose. 
Instead you seem to post random changes to your code and ask new 
questions which makes it hard to know whether you understand what is 
going on and to guide you to a solution.



Then,  maxval determines the max value on line 11:
if val  maxval:

right, view a copy of the revised code at
http://tinyurl.com/nvzdw8k


OK, You have only posted partial code because you don't have
the part where you read the filename etc. But assuming it
has not changed then the new code looks correct - almost.

Your assignment said NOT to use the lines starting From: (with
colon :) But your code tests for startswith('From: ') so you
are doing what you were told not to do.

If you remove the colon from the end of From (but keep the space)
it should be correct.


Question1: are these assumptions true, above?


Yes, pretty much correct.


Question2: dict maps strings into keys and values?


Not quite. A dictionary has a key and a corresponding value.
The key can be ANY immutable (unchangeable) object - string,
integer, tuple, etc.
The value can be any Python object.


--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

Mark,

Visit the following URL link to view a captured copy of the latest code
revision, available at http://tinyurl.com/nvzdw8k

Regards,
Hal

On Thu, Aug 6, 2015 at 10:17 AM, Ltc Hotspot ltc.hots...@gmail.com wrote:



 On Thu, Aug 6, 2015 at 8:28 AM, Mark Lawrence breamore...@yahoo.co.uk
 wrote:

 On 06/08/2015 05:22, Ltc Hotspot wrote:

 Please don't top post here, it makes following long threads difficult.

 Mark,

 Replace  count[address]= count.get(address,0) +1 with  c =
 Counter(['address'])?


 Try it at the interactive prompt and see what happens.

 How do I define counter,view trace back:

 NameError
 Traceback (most recent call last)
 C:\Users\vm\Desktop\apps\docs\Python\new.txt in module()
   1 fname = raw_input(Enter file name: )
   2 handle = open (fname, 'r')
  3 c = Counter(['address'])
   4
   5

 NameError: name 'Counter' is not defined


 View revised code here:

 fname = raw_input(Enter file name: )
 handle = open (fname, 'r')
 c = Counter(['address'])

 count = dict ()
 maxval = None
 maxkee = None

 for kee, val in count.items():
 maxval = val
 maxkee = kee

 for line in handle:
 if line.startswith(From: ):
 address = line.split()[1]
 count[address]= count.get(address,0) +1
 print maxkee and maxval


 In [20]:
 Hal





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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

Hi Alan,

I moved counter outside the loop and below dict, maxval = None
maxkee = None are both positioned outside the loop.

URL link to the revisions are available at http://tinyurl.com/nvzdw8k

Question: How do I define Counter

Revised code reads:
fname = raw_input(Enter file name: )
handle = open (fname, 'r')

counter = dict ()
c = Counter(['address'])

maxval = None
maxkee = None

for line in handle:
if line.startswith(From: ):
address = line.split()[1]

for maxkee, val in c.items():

maxval = val
maxkee = kee

print maxkee and maxval


Traceback message reads:
NameError
Traceback (most recent call last)
C:\Users\vm\Desktop\apps\docs\Python\assignment_9_4_16.py in module()
  3
  4 counter = dict ()
 5 c = Counter(['address'])
  6
  7 maxval = None

NameError: name 'Counter' is not defined


Regards,
Hal

On Thu, Aug 6, 2015 at 2:47 AM, Alan Gauld alan.ga...@btinternet.com
wrote:

 On 06/08/15 03:27, Ltc Hotspot wrote:

 The output as reported by the latest code revision: c...@iupui.edu
 mailto:c...@iupui.edu 1← Mismatch

 Looks like python continues to print the wrong data set:


 Python will print what you ask it to. Don't blame the tool!  :-)

  for line in handle:
   if line.startswith(From: ):
   address = line.split()[1]
   count[address]= count.get(address,0) +1
 
  maxval = None
  maxkee = None
  for kee, val in count.items():
 
  maxval = val
  maxkee = kee
 
  print address, val

 Look at the loops.

 In the second loop you are no longer setting the values to
 those of the max item but are setting them every time.
 So at the end of the loop val holds the val of
 the last item (and so does maxval so even if you used
 that it would be the same result).

 Similarly with the code for address. You are setting that
 for each 'From ' line in your file so at the end of the loop
 address is the last address in the file.

 Now, dictionaries do not store data in the order that you
 insert it, so there is no guarantee that the last item in
 the dictionary loop is the same as the last address
 you read.

 You need to reinstate the test for max val in the second
 loop and then print the kee that corresponds with that
 (maxkee) as the address. ie. print maxkee and maxval.


 --
 Alan G
 Author of the Learn to Program web site
 http://www.alan-g.me.uk/
 http://www.amazon.com/author/alan_gauld
 Follow my photo-blog on Flickr at:
 http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 06/08/15 19:30, Ltc Hotspot wrote:


I moved counter outside the loop and below dict, maxval = None
maxkee = None are both positioned outside the loop.


You moved counter but it is still a dict() and you
don't use it anywhere.


URL link to the revisions are available at http://tinyurl.com/nvzdw8k

Question: How do I define Counter


Counter is defined for you in the collections module.
So to use it you need to import collections and access it as 
collections.Counter.


But did you read how to use it? It is a lot more than
just a dictionary, it has many extra methods, some of
which almost solve your problem for you. (Whether your
teacher will approve of using Counter is another
issue!)


Revised code reads:
fname = raw_input(Enter file name: )
handle = open (fname, 'r')

counter = dict ()
c = Counter(['address'])


You only need to pass a list if you are adding multiple things.

But by the same token you can add a list of items, such
as email addresses. So if you had such a list you could
create a Counter() to hold them and count them for you.
And return the one with the highest value.
Sound familiar?

Please (re)read the Counter documentation.
Then play with one in the  prompt.
Don't expect us to just provide you with code, learn
how it works for yourself. Experiment.

The  prompt is your friend. You will learn more from that in 15 
minutes than in a bunch of emails showing other peoples

code.

Alternatively forget about Counter and just go back to
your dict(). You have written all the code you need already,
you just need to assemble it in the correct order.


maxval = None
maxkee = None

for line in handle:
 if line.startswith(From: ):
 address = line.split()[1]


You are not storing the addresses anywhere.


for maxkee, val in c.items():

 maxval = val
 maxkee = kee


You are still not testing if its the maximum,
you just keep overwriting the variables for
each element.


print maxkee and maxval


You still have an 'and' in there.

--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 06/08/15 03:27, Ltc Hotspot wrote:
The output as reported by the latest code revision: c...@iupui.edu 
mailto:c...@iupui.edu 1← Mismatch


Looks like python continues to print the wrong data set:


Python will print what you ask it to. Don't blame the tool!  :-)

 for line in handle:
  if line.startswith(From: ):
  address = line.split()[1]
  count[address]= count.get(address,0) +1

 maxval = None
 maxkee = None
 for kee, val in count.items():

 maxval = val
 maxkee = kee

 print address, val

Look at the loops.

In the second loop you are no longer setting the values to
those of the max item but are setting them every time.
So at the end of the loop val holds the val of
the last item (and so does maxval so even if you used
that it would be the same result).

Similarly with the code for address. You are setting that
for each 'From ' line in your file so at the end of the loop
address is the last address in the file.

Now, dictionaries do not store data in the order that you
insert it, so there is no guarantee that the last item in
the dictionary loop is the same as the last address
you read.

You need to reinstate the test for max val in the second
loop and then print the kee that corresponds with that
(maxkee) as the address. ie. print maxkee and maxval.

--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos

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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

Mark,

Replace  count[address]= count.get(address,0) +1 with  c =
Counter(['address'])?

Regards,
Hal

On Wed, Aug 5, 2015 at 9:03 PM, Mark Lawrence breamore...@yahoo.co.uk
wrote:

 On 05/08/2015 23:58, Ltc Hotspot wrote:

 Hi Mark,

 Address identifies the email address with the maximum  number of sends:
 c...@iupui.edu.

 Secondly, we are missing a count on the number of messages sent by
 c...@iupui.edu, i.e., 5.

 Thirdly, maxval 'none' is not defined on line # 24

 Questions: How do we define the value of none for the key maxval and
 retrieve a number count on the number of messages sent by c...@iupui.edu.


 NameError:

 Traceback (most recent call last)
 C:\Users\vm\Desktop\apps\docs\Python\assignment_9_4_5.py in module()
   22 ## find the greatest number of mail messages.
   23
 --- 24 maxval = none
   25 maxkee = none
   26 for kee, val in count.items():

 NameError: name 'none' is not defined

 In [52]: print address
 c...@iupui.edu

 Revised data:


 ## The program looks for 'From ' lines and takes the second
 ## word of those lines as the person who sent the mail.

 fname = raw_input(Enter file name: )
 handle = open (fname, 'r')
 for line in handle:
  if line.startswith(From: ):
  address = line.split()[1]


 ## The program creates a Python dictionary that maps
 ## the sender's mail address to a count of the number
 ## of times they appear in the file.

  count = dict()
  for wrd in address:
  count[wrd]= count.get(wrd,0) +1

 ## After the dictionary is produced, the program reads
 ## through the dictionary using a maximum loop to
 ## find the greatest number of mail messages.

 maxval = none
 maxkee = none
 for kee, val in count.items():
  if maxval == none or maxval val:
  maxval = val
  maxkee = kee


 You can greatly simplify all of the above code if you use a Counter from
 the collections module
 https://docs.python.org/3/library/collections.html#collections.Counter

 --
 My fellow Pythonistas, ask not what our language can do for you, ask
 what you can do for our language.

 Mark Lawrence

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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

On my breath and soul, I did:

Counter objects have a dictionary interface except that they return a zero
count for missing items instead of raising a KeyError
https://docs.python.org/3/library/exceptions.html#KeyError:


 c = Counter(['eggs', 'ham'])


On Thu, Aug 6, 2015 at 11:59 AM, Mark Lawrence breamore...@yahoo.co.uk
wrote:

 On 06/08/2015 18:17, Ltc Hotspot wrote:

 On Thu, Aug 6, 2015 at 8:28 AM, Mark Lawrence breamore...@yahoo.co.uk
 wrote:

 On 06/08/2015 05:22, Ltc Hotspot wrote:

 Please don't top post here, it makes following long threads difficult.

 Mark,


 Replace  count[address]= count.get(address,0) +1 with  c =
 Counter(['address'])?


 Try it at the interactive prompt and see what happens.

 How do I define counter,view trace back:

 NameError
 Traceback (most recent call last)
 C:\Users\vm\Desktop\apps\docs\Python\new.txt in module()
1 fname = raw_input(Enter file name: )
2 handle = open (fname, 'r')
  3 c = Counter(['address'])
4
5

 NameError: name 'Counter' is not defined


 View revised code here:

 fname = raw_input(Enter file name: )
 handle = open (fname, 'r')
 c = Counter(['address'])

 count = dict ()
 maxval = None
 maxkee = None

 for kee, val in count.items():
  maxval = val
  maxkee = kee

 for line in handle:
  if line.startswith(From: ):
  address = line.split()[1]
  count[address]= count.get(address,0) +1
 print maxkee and maxval


 You obviously haven't bothered to read the link I gave you about the
 Counter class so I give up.

 --
 My fellow Pythonistas, ask not what our language can do for you, ask
 what you can do for our language.

 Mark Lawrence

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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 07/08/15 00:11, Ltc Hotspot wrote:
Questions(1):Why does print line, prints blank space; and, (2) print 
address prints a single email address:


See my previous emails.
You are not storing your addresses so address only holds the last 
address in the file.

line is at the end of the file so is empty.,


In [72]: print count
{'gopal.ramasammyc...@gmail.com 
mailto:gopal.ramasammyc...@gmail.com': 1, 'lo...@media.berkeley.edu 
mailto:lo...@media.berkeley.edu': 3, 'cwen@iupui.
edu': 5, 'antra...@caret.cam.ac.uk mailto:antra...@caret.cam.ac.uk': 
1, 'rjl...@iupui.edu mailto:rjl...@iupui.edu': 2, 'gsil...@umich.ed
u': 3, 'david.horw...@uct.ac.za mailto:david.horw...@uct.ac.za': 4, 
'wagne...@iupui.edu mailto:wagne...@iupui.edu': 1, 'zq...@umich.edu 
mailto:zq...@umich.edu':
 4, 'stephen.marqu...@uct.ac.za mailto:stephen.marqu...@uct.ac.za': 
2, 'r...@media.berkeley.edu mailto:r...@media.berkeley.edu': 1}


Question(3): why did the elements print count('keys') and print 
count('items') fail?


Because, as shown above, count is a dictionary.
So items and keys are methods not strings to be passed
to a non-existent count() function.

So you need, for example:

print count.keys()


Traceback (most recent call last)
ipython-input-76-35c8707b256e in module()
 1 print count('items')

TypeError: 'dict' object is not callable



Which is what the error is also telling you.
You cannot call - ie use () - with a dictionary like count.

--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos

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Re: [Tutor] Dictionary Issue

[Mark Lawrence]

On 06/08/2015 20:05, Ltc Hotspot wrote:

On my breath and soul, I did:

Counter objects have a dictionary interface except that they return a zero
count for missing items instead of raising a KeyError
https://docs.python.org/3/library/exceptions.html#KeyError:


That's nice to know.  What do the rest of the methods on the class do?


Please don't top post here, it makes following long threads difficult.


What did you not understand about the above?


You obviously haven't bothered to read the link I gave you about the
Counter class so I give up.



If you'd read the entire write up why are you still wasting time with a 
loop to find a maximum that simply doesn't work, when there is likely a 
solution in the Counter class right in front of your eyes?


--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence

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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

On Thu, Aug 6, 2015 at 3:00 PM, Alan Gauld alan.ga...@btinternet.com
wrote:

 On 06/08/15 19:30, Ltc Hotspot wrote:

 I moved counter outside the loop and below dict, maxval = None
 maxkee = None are both positioned outside the loop.


 You moved counter but it is still a dict() and you
 don't use it anywhere.

 URL link to the revisions are available at http://tinyurl.com/nvzdw8k

 Question: How do I define Counter


 Counter is defined for you in the collections module.
 So to use it you need to import collections and access it as
 collections.Counter.

 But did you read how to use it? It is a lot more than
 just a dictionary, it has many extra methods, some of
 which almost solve your problem for you. (Whether your
 teacher will approve of using Counter is another
 issue!)

 Revised code reads:
 fname = raw_input(Enter file name: )
 handle = open (fname, 'r')

 counter = dict ()
 c = Counter(['address'])


 You only need to pass a list if you are adding multiple things.

 But by the same token you can add a list of items, such
 as email addresses. So if you had such a list you could
 create a Counter() to hold them and count them for you.
 And return the one with the highest value.
 Sound familiar?

 Please (re)read the Counter documentation.
 Then play with one in the  prompt.
 Don't expect us to just provide you with code, learn
 how it works for yourself. Experiment.

 The  prompt is your friend. You will learn more from that in 15 minutes
 than in a bunch of emails showing other peoples
 code.

 Alternatively forget about Counter and just go back to
 your dict(). You have written all the code you need already,
 you just need to assemble it in the correct order.

 maxval = None
 maxkee = None

 for line in handle:
  if line.startswith(From: ):
  address = line.split()[1]


 You are not storing the addresses anywhere.

 for maxkee, val in c.items():

  maxval = val
  maxkee = kee


 You are still not testing if its the maximum,
 you just keep overwriting the variables for
 each element.

 print maxkee and maxval


 You still have an 'and' in there.

 --
 Alan G
 Author of the Learn to Program web site
 http://www.alan-g.me.uk/
 http://www.amazon.com/author/alan_gauld
 Follow my photo-blog on Flickr at:
 http://www.flickr.com/photos/alangauldphotos


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Hi Alan,

Questions(1):Why does print line, prints blank space; and, (2) print
address prints a single email address:

View print results as follows:

In [70]: %run assignment_9_4_24.py
Enter file name: mbox-short.txt
r...@media.berkeley.edu 1

In [71]: print handle
open file 'mbox-short.txt', mode 'r' at 0x035576F0

In [72]: print count
{'gopal.ramasammyc...@gmail.com': 1, 'lo...@media.berkeley.edu': 3,
'cwen@iupui.
edu': 5, 'antra...@caret.cam.ac.uk': 1, 'rjl...@iupui.edu': 2,
'gsil...@umich.ed
u': 3, 'david.horw...@uct.ac.za': 4, 'wagne...@iupui.edu': 1, '
zq...@umich.edu':
 4, 'stephen.marqu...@uct.ac.za': 2, 'r...@media.berkeley.edu': 1}

In [73]: print line


In [74]: print address
c...@iupui.edu


Question(3): why did the elements print count('keys') and print
count('items') fail?

View print commands as follows:


In [75]: dir (count)
Out[75]:
['__class__',
 '__cmp__',
 '__contains__',
 '__delattr__',
 '__delitem__',
 '__doc__',
 '__eq__',
 '__format__',
 '__ge__',
 '__getattribute__',
 '__getitem__',
 '__gt__',
 '__hash__',
 '__init__',
 '__iter__',
 '__le__',
 '__len__',
 '__lt__',
 '__ne__',
 '__new__',
 '__reduce__',
 '__reduce_ex__',
 '__repr__',
 '__setattr__',
 '__setitem__',
 '__sizeof__',
 '__str__',
 '__subclasshook__',
 'clear',
 'copy',
 'fromkeys',
 'get',
 'has_key',
 'items',
 'iteritems',
 'iterkeys',
 'itervalues',
 'keys',
 'pop',
 'popitem',
 'setdefault',
 'update',
 'values',
 'viewitems',
 'viewkeys',
 'viewvalues']

In [76]:

---
TypeError
Traceback (most recent call last)
ipython-input-76-35c8707b256e in module()
 1 print count('items')

TypeError: 'dict' object is not callable

In [77]: print count('keys')
---
TypeError
Traceback (most recent call last)
ipython-input-77-54ed4a05a3c7 in module()
 1 print count('keys')

TypeError: 'dict' object is not callable

In [78]:


Regards,
Hal
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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 06/08/15 18:17, Ltc Hotspot wrote:


View revised code here:

fname = raw_input(Enter file name: )
handle = open (fname, 'r')
c = Counter(['address'])

count = dict ()
maxval = None
maxkee = None

for kee, val in count.items():
 maxval = val
 maxkee = kee

for line in handle:
 if line.startswith(From: ):
 address = line.split()[1]
 count[address]= count.get(address,0) +1
print maxkee and maxval


Let's hold up here a second. You do understand that
Python executes your code from top to bottom, yes?

So reading your code from the top you have a loop that
sets maxval and maxkee before you even put anything
into count. How do you think that would ever work?

You have a lot of people spending time trying to
help you here, but you do need to exercise a little
bit of insight yourself. That code can never work
and it has nothing to do with Pyhon it is your
logic that is faulty.

Try working through it with a pencil and paper.
Write down what each variable contains at each
stage of the program. (or just print it to see).

You have been very close to a solution but you
seem to be getting farther away rather than closer
which suggests you are trying stuff without
understanding why.

--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

On Thu, Aug 6, 2015 at 8:28 AM, Mark Lawrence breamore...@yahoo.co.uk
wrote:

 On 06/08/2015 05:22, Ltc Hotspot wrote:

 Please don't top post here, it makes following long threads difficult.

 Mark,

 Replace  count[address]= count.get(address,0) +1 with  c =
 Counter(['address'])?


 Try it at the interactive prompt and see what happens.

 How do I define counter,view trace back:

 NameError
 Traceback (most recent call last)
 C:\Users\vm\Desktop\apps\docs\Python\new.txt in module()
   1 fname = raw_input(Enter file name: )
   2 handle = open (fname, 'r')
  3 c = Counter(['address'])
   4
   5

NameError: name 'Counter' is not defined


View revised code here:

fname = raw_input(Enter file name: )
handle = open (fname, 'r')
c = Counter(['address'])

count = dict ()
maxval = None
maxkee = None

for kee, val in count.items():
maxval = val
maxkee = kee

for line in handle:
if line.startswith(From: ):
address = line.split()[1]
count[address]= count.get(address,0) +1
print maxkee and maxval


In [20]:
Hal





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Re: [Tutor] Dictionary Issue

[Mark Lawrence]

On 06/08/2015 18:17, Ltc Hotspot wrote:

On Thu, Aug 6, 2015 at 8:28 AM, Mark Lawrence breamore...@yahoo.co.uk
wrote:


On 06/08/2015 05:22, Ltc Hotspot wrote:

Please don't top post here, it makes following long threads difficult.

Mark,


Replace  count[address]= count.get(address,0) +1 with  c =
Counter(['address'])?



Try it at the interactive prompt and see what happens.

How do I define counter,view trace back:

NameError
Traceback (most recent call last)
C:\Users\vm\Desktop\apps\docs\Python\new.txt in module()
   1 fname = raw_input(Enter file name: )
   2 handle = open (fname, 'r')
 3 c = Counter(['address'])
   4
   5


NameError: name 'Counter' is not defined


View revised code here:

fname = raw_input(Enter file name: )
handle = open (fname, 'r')
c = Counter(['address'])

count = dict ()
maxval = None
maxkee = None

for kee, val in count.items():
 maxval = val
 maxkee = kee

for line in handle:
 if line.startswith(From: ):
 address = line.split()[1]
 count[address]= count.get(address,0) +1
print maxkee and maxval



You obviously haven't bothered to read the link I gave you about the 
Counter class so I give up.


--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence

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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 06/08/15 18:17, Ltc Hotspot wrote:


Replace  count[address]= count.get(address,0) +1 with  c =
Counter(['address'])?


You didn't do what Mark suggested.
Did you read the documentation about Counter?


NameError
Traceback (most recent call last)
   2 handle = open (fname, 'r')
 3 c = Counter(['address'])


NameError: name 'Counter' is not defined


So where do you define Counter? Do you import the module?
Do you import Counter from the module?
It's not shown in your code.


View revised code here:

fname = raw_input(Enter file name: )
handle = open (fname, 'r')
c = Counter(['address'])

count = dict ()
maxval = None
maxkee = None

for kee, val in count.items():
 maxval = val
 maxkee = kee

for line in handle:
 if line.startswith(From: ):
 address = line.split()[1]
 count[address]= count.get(address,0) +1
print maxkee and maxval


You have introduced an 'and' here which makes no sense.
It will try to print the logical AND of the two values.
That's not what you want.

Please try to work through this manually and see
how it works (or doesn't). There is no point in folks
making suggestions for improvements until you
understand how it should work yourself.b You have
all the components needed to build the solution,
now its up to you to fit them together such that
they work. We can make suggestions but you need
to solve the problem, we can't, and won't do it
for you.

--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

The revised output reads:

In [3]: %run assignment_9_4_9.py
Enter file name: mbox-short.txt
c...@iupui.edu 14

The desired output: c...@iupui.edu 5


Question: How do I trace the source of the count?

Revised data code reads:

fname = raw_input(Enter file name: )
handle = open (fname, 'r')
count = dict ()
for line in handle:
if line.startswith(From: ):
address = line.split()[1]

for wrd in address:
count[wrd]= count.get(wrd,0) +1

maxval = None
maxkee = None
for kee, val in count.items():

maxval = val
maxkee = kee

print address, val





On Wed, Aug 5, 2015 at 4:11 PM, Alan Gauld alan.ga...@btinternet.com
wrote:

 On 05/08/15 15:15, Ltc Hotspot wrote:

 Raw data code reads:


 Being picky here but data and code are very different
 things (in most languages at least) and what you have
 below is definitely code not data.

 Meanwhile there are lots of issues in this code...

 fname = raw_input(Enter file name: )
 handle = open (fname, 'r')
 text = handle.read()

 ## The program looks for 'From ' lines and takes the second
 ## word of those lines as the person who sent the mail.

 addresses = set()
 for addr in [ text.split()[2]('From  ')
  if fromline


 The above looks like its supposed to be a list
 comprehension embedded in a for loop. Putting too much
 code in one line is usually a bad idea especially before
 you have it working.

 Try separating out the formation of your list from the
 for loop. Once you get the comprehension working correctly
 then you can consider embedding it.

 As for the expression

 text.split()[2]('From  ')

 Can you explain how you think that works?
 Try it at the  prompt with text set to
 a sample line of data.


-- What command did you type to get the triple chevrons ?

-- My python interpreter:  iPython (py.2.7)


 Try

  text = .. # whatever your data looks like
  text.split()

  text.split[2]

  text.split()[2]('From  ')


--  address data, review the latest revised code?



 The  prompt is one of your most powerful tools while
 writing code, you should always have one ready to try
 stuff out. You can answer a lot of questions that way.

 ## The program creates a Python dictionary that maps
 ## the sender's mail address to a count of the number
 ## of times they appear in the file.

  count = dict()
  for wrd in word:


 What is word? You don't define it anywhere?

  count[wrd]= count.get(wrd,0) +1

 ## After the dictionary is produced, the program reads
 ## through the dictionary using a maximum loop to


 -- imported address data, review revised code?






 --
 Alan G
 Author of the Learn to Program web site
 http://www.alan-g.me.uk/
 http://www.amazon.com/author/alan_gauld
 Follow my photo-blog on Flickr at:
 http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

Hi Alan


The output as reported by the latest code revision: c...@iupui.edu 1 ←
Mismatch

Looks like python continues to print the wrong data set:

In [11]: print val
1 In [12]: print kee r...@media.berkeley.edu In [13]: print address
c...@iupui.edu
In order to complete the assignment, using data from the source  file,
 python must print the email address of the maximum sender and the number
of sends, i.e.,  c...@iupui.edu 5

I think the problem is in the placement of the counter?

Question: What is the source of the dictionary keys and values:
maxval = None
maxkee = None

Here is the latest revised code as follows:

fname = raw_input(Enter file name: )
handle = open (fname, 'r')
count = dict ()
for line in handle:
if line.startswith(From: ):
address = line.split()[1]
count[address]= count.get(address,0) +1

maxval = None
maxkee = None
for kee, val in count.items():

maxval = val
maxkee = kee

print address, val





Hal




On Wed, Aug 5, 2015 at 6:21 PM, Alan Gauld alan.ga...@btinternet.com
wrote:

 On 05/08/15 23:58, Ltc Hotspot wrote:

 fname = raw_input(Enter file name: )
 handle = open (fname, 'r')
 for line in handle:
  if line.startswith(From: ):
  address = line.split()[1]


 So far so good.


 ## The program creates a Python dictionary that maps
 ## the sender's mail address to a count of the number
 ## of times they appear in the file.

  count = dict()


 But here you create a brand new dictionary.
 Every time you go round the loop.
 And it wipes out the old one.
 You need to move that out of the loop.

  for wrd in address:


 address is a string. So wrd will be set to every
 character in the string. I don;t think that's what
 you want?

  count[wrd]= count.get(wrd,0) +1

 ## After the dictionary is produced, the program reads
 ## through the dictionary using a maximum loop to
 ## find the greatest number of mail messages.

 maxval = none
 maxkee = none


 See my previous email. none should be None.
 Case matters in Python.

 for kee, val in count.items():
  if maxval == none or maxval val:
  maxval = val
  maxkee = kee


 #items are printed

 print address


 Notice that address gets reset every time the loop reads
 a new line so this will only print the last address.
 But maybe that's what you wanted?

 -- Did I resolve the reset  in the revised code?



 --
 Alan G
 Author of the Learn to Program web site
 http://www.alan-g.me.uk/
 http://www.amazon.com/author/alan_gauld
 Follow my photo-blog on Flickr at:
 http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Mark Lawrence]

On 06/08/2015 05:22, Ltc Hotspot wrote:

Please don't top post here, it makes following long threads difficult.


Mark,

Replace  count[address]= count.get(address,0) +1 with  c =
Counter(['address'])?


Try it at the interactive prompt and see what happens.



Regards,
Hal

On Wed, Aug 5, 2015 at 9:03 PM, Mark Lawrence breamore...@yahoo.co.uk
wrote:


On 05/08/2015 23:58, Ltc Hotspot wrote:


Hi Mark,

Address identifies the email address with the maximum  number of sends:
c...@iupui.edu.

Secondly, we are missing a count on the number of messages sent by
c...@iupui.edu, i.e., 5.

Thirdly, maxval 'none' is not defined on line # 24

Questions: How do we define the value of none for the key maxval and
retrieve a number count on the number of messages sent by c...@iupui.edu.


NameError:

Traceback (most recent call last)
C:\Users\vm\Desktop\apps\docs\Python\assignment_9_4_5.py in module()
   22 ## find the greatest number of mail messages.
   23
--- 24 maxval = none
   25 maxkee = none
   26 for kee, val in count.items():

NameError: name 'none' is not defined

In [52]: print address
c...@iupui.edu

Revised data:


## The program looks for 'From ' lines and takes the second
## word of those lines as the person who sent the mail.

fname = raw_input(Enter file name: )
handle = open (fname, 'r')
for line in handle:
  if line.startswith(From: ):
  address = line.split()[1]


## The program creates a Python dictionary that maps
## the sender's mail address to a count of the number
## of times they appear in the file.

  count = dict()
  for wrd in address:
  count[wrd]= count.get(wrd,0) +1

## After the dictionary is produced, the program reads
## through the dictionary using a maximum loop to
## find the greatest number of mail messages.

maxval = none
maxkee = none
for kee, val in count.items():
  if maxval == none or maxval val:
  maxval = val
  maxkee = kee



You can greatly simplify all of the above code if you use a Counter from
the collections module
https://docs.python.org/3/library/collections.html#collections.Counter

--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence



--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence

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Re: [Tutor] Dictionary Issue

[Mark Lawrence]

On 05/08/2015 23:58, Ltc Hotspot wrote:

Hi Mark,

Address identifies the email address with the maximum  number of sends:
c...@iupui.edu.

Secondly, we are missing a count on the number of messages sent by
c...@iupui.edu, i.e., 5.

Thirdly, maxval 'none' is not defined on line # 24

Questions: How do we define the value of none for the key maxval and
retrieve a number count on the number of messages sent by c...@iupui.edu.


NameError:

Traceback (most recent call last)
C:\Users\vm\Desktop\apps\docs\Python\assignment_9_4_5.py in module()
  22 ## find the greatest number of mail messages.
  23
--- 24 maxval = none
  25 maxkee = none
  26 for kee, val in count.items():

NameError: name 'none' is not defined

In [52]: print address
c...@iupui.edu

Revised data:


## The program looks for 'From ' lines and takes the second
## word of those lines as the person who sent the mail.

fname = raw_input(Enter file name: )
handle = open (fname, 'r')
for line in handle:
 if line.startswith(From: ):
 address = line.split()[1]


## The program creates a Python dictionary that maps
## the sender's mail address to a count of the number
## of times they appear in the file.

 count = dict()
 for wrd in address:
 count[wrd]= count.get(wrd,0) +1

## After the dictionary is produced, the program reads
## through the dictionary using a maximum loop to
## find the greatest number of mail messages.

maxval = none
maxkee = none
for kee, val in count.items():
 if maxval == none or maxval val:
 maxval = val
 maxkee = kee



You can greatly simplify all of the above code if you use a Counter from 
the collections module 
https://docs.python.org/3/library/collections.html#collections.Counter


--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence

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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 06/08/15 02:05, Ltc Hotspot wrote:

The revised output reads:

In [3]: %run assignment_9_4_9.py
Enter file name: mbox-short.txt
c...@iupui.edu mailto:c...@iupui.edu 14

The desired output: c...@iupui.edu mailto:c...@iupui.edu 5



See my other post.
Count the number of letters in the address.


--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos

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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 05/08/15 23:58, Ltc Hotspot wrote:


fname = raw_input(Enter file name: )
handle = open (fname, 'r')
for line in handle:
 if line.startswith(From: ):
 address = line.split()[1]



So far so good.



## The program creates a Python dictionary that maps
## the sender's mail address to a count of the number
## of times they appear in the file.

 count = dict()


But here you create a brand new dictionary.
Every time you go round the loop.
And it wipes out the old one.
You need to move that out of the loop.


 for wrd in address:


address is a string. So wrd will be set to every
character in the string. I don;t think that's what
you want?


 count[wrd]= count.get(wrd,0) +1

## After the dictionary is produced, the program reads
## through the dictionary using a maximum loop to
## find the greatest number of mail messages.

maxval = none
maxkee = none


See my previous email. none should be None.
Case matters in Python.


for kee, val in count.items():
 if maxval == none or maxval val:
 maxval = val
 maxkee = kee


#items are printed

print address


Notice that address gets reset every time the loop reads
a new line so this will only print the last address.
But maybe that's what you wanted?



--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 05/08/15 15:15, Ltc Hotspot wrote:


Raw data code reads:


Being picky here but data and code are very different
things (in most languages at least) and what you have
below is definitely code not data.

Meanwhile there are lots of issues in this code...


fname = raw_input(Enter file name: )
handle = open (fname, 'r')
text = handle.read()

## The program looks for 'From ' lines and takes the second
## word of those lines as the person who sent the mail.

addresses = set()
for addr in [ text.split()[2]('From  ')
 if fromline


The above looks like its supposed to be a list
comprehension embedded in a for loop. Putting too much
code in one line is usually a bad idea especially before
you have it working.

Try separating out the formation of your list from the
for loop. Once you get the comprehension working correctly
then you can consider embedding it.

As for the expression

text.split()[2]('From  ')

Can you explain how you think that works?
Try it at the  prompt with text set to
a sample line of data.

Try

 text = .. # whatever your data looks like
 text.split()

 text.split[2]

 text.split()[2]('From  ')

The  prompt is one of your most powerful tools while
writing code, you should always have one ready to try
stuff out. You can answer a lot of questions that way.


## The program creates a Python dictionary that maps
## the sender's mail address to a count of the number
## of times they appear in the file.

 count = dict()
 for wrd in word:


What is word? You don't define it anywhere?


 count[wrd]= count.get(wrd,0) +1

## After the dictionary is produced, the program reads
## through the dictionary using a maximum loop to
## find the greatest number of mail messages.

maxval = none
maxkee = none


none is not a value in Python.
You need to spell it None


for kee, val in count.items():
 if maxval == none or maxval val:


You don't really need the maxval == None check since None
is considered less that all other numbers.


 maxval = val
 maxkee = kee

print items


What is items? It is not defined anywhere above.
count.items is not the same as items.


--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Mark Lawrence]

On 05/08/2015 15:15, Ltc Hotspot wrote:

Hi everyone:

I want to write a python program that reads through the data file of
mbox-short.txt.Mbox-short.txt, i.e.,  download is available at
http://www.py4inf.com/code/mbox-short.txt.

Secondly, I want for python to figure out who sent the greatest number of
mail messages.

The output should read: c...@iupui.edu 5

However, there is a traceback message:

In [40]: %run 9_4_4.py
   File C:\Users\vm\Desktop\apps\docs\Python\_9_4_4.py, line 19
 count = dict()
 ^
SyntaxError: invalid syntax

Raw data code reads:

fname = raw_input(Enter file name: )
handle = open (fname, 'r')
text = handle.read()

## The program looks for 'From ' lines and takes the second
## word of those lines as the person who sent the mail.

addresses = set()
for addr in [ text.split()[2]('From  ')
 if fromline

## The program creates a Python dictionary that maps
## the sender's mail address to a count of the number
## of times they appear in the file.

 count = dict()
 for wrd in word:
 count[wrd]= count.get(wrd,0) +1

## After the dictionary is produced, the program reads
## through the dictionary using a maximum loop to
## find the greatest number of mail messages.

maxval = none
maxkee = none
for kee, val in count.items():
 if maxval == none or maxval val:
 maxval = val
 maxkee = kee

print items

Question: What is the cause of the dictionary line message?



Learning to find these syntax errors is a skill you must learn for 
yourself.  Very often the error occurs one or even more lines before the 
line on which the error is actually detected.  So please go back from 
where the error is actually reported and see what you can find.  Very 
strong hint, do not ask again until you've fixed at least two of the 
three errors, and there may be more :)


--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence

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Re: [Tutor] Dictionary Issue

[Clayton Kirkwood]

 -Original Message-
 From: Tutor [mailto:tutor-bounces+crk=godblessthe...@python.org] On
 Behalf Of Mark Lawrence
 Sent: Wednesday, August 05, 2015 3:23 PM
 To: tutor@python.org
 Subject: Re: [Tutor] Dictionary Issue
 
 On 05/08/2015 15:15, Ltc Hotspot wrote:
  Hi everyone:
 
  I want to write a python program that reads through the data file of
  mbox-short.txt.Mbox-short.txt, i.e.,  download is available at
  http://www.py4inf.com/code/mbox-short.txt.
 
  Secondly, I want for python to figure out who sent the greatest number
  of mail messages.
 
  The output should read: c...@iupui.edu 5
 
  However, there is a traceback message:
 
  In [40]: %run 9_4_4.py
 File C:\Users\vm\Desktop\apps\docs\Python\_9_4_4.py, line 19
   count = dict()
   ^
  SyntaxError: invalid syntax
 
  Raw data code reads:
 
  fname = raw_input(Enter file name: ) handle = open (fname, 'r') text
  = handle.read()
 
  ## The program looks for 'From ' lines and takes the second ## word of
  those lines as the person who sent the mail.
 
  addresses = set()
  for addr in [ text.split()[2]('From  ')
   if fromline
 
  ## The program creates a Python dictionary that maps ## the sender's
  mail address to a count of the number ## of times they appear in the
  file.
 
   count = dict()
   for wrd in word:
   count[wrd]= count.get(wrd,0) +1
 
  ## After the dictionary is produced, the program reads ## through the
  dictionary using a maximum loop to ## find the greatest number of mail
  messages.
 
  maxval = none
  maxkee = none
  for kee, val in count.items():
   if maxval == none or maxval val:
   maxval = val
   maxkee = kee
 
  print items
 
  Question: What is the cause of the dictionary line message?
 
 
 Learning to find these syntax errors is a skill you must learn for
yourself.  Very
 often the error occurs one or even more lines before the line on which the
 error is actually detected.  So please go back from where the error is
actually
 reported and see what you can find.  Very strong hint, do not ask again
until
 you've fixed at least two of the three errors, and there may be more :)
 

It looks like the problem is with count=dict()
Should be count=dict{}

I may be wrong - U'm still a neophyte.
crk

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Re: [Tutor] Dictionary Issue

[Alan Gauld]

On 05/08/15 23:36, Clayton Kirkwood wrote:


It looks like the problem is with count=dict()
Should be count=dict{}

I may be wrong - U'm still a neophyte.


Yes, you're wrong! :-)

the correct form is as shown

count = dict()

Its calling the type operation which looks like
any other function and uses ().

The more common way to create an empty dictionary
is to use {} alone, like:

count = {}

Although count is probably a bad name choice since
collections usually merit a plural name, eg counts

But the OP's problems lie earlier in the code,
as Mark has suggested.

--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] Dictionary Issue

[Danny Yoo]

 However, there is a traceback message:

 In [40]: %run 9_4_4.py
   File C:\Users\vm\Desktop\apps\docs\Python\_9_4_4.py, line 19
 count = dict()
 ^
 SyntaxError: invalid syntax

Syntax error reporting is approximate: you might need to look a few lines
earlier to get at the root problem.

... Reading...

The for loop looks odd because there's a leading open bracket that looks
out of place.

for addr in [ text.split()[2]('From  ')

The following line uses an if conditional, but needs to offset the block
with a : that appears to be missing.

if fromline
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Re: [Tutor] Dictionary Issue

[Ltc Hotspot]

Hi Mark,

Address identifies the email address with the maximum  number of sends:
c...@iupui.edu.

Secondly, we are missing a count on the number of messages sent by
c...@iupui.edu, i.e., 5.

Thirdly, maxval 'none' is not defined on line # 24

Questions: How do we define the value of none for the key maxval and
retrieve a number count on the number of messages sent by c...@iupui.edu.


NameError:

Traceback (most recent call last)
C:\Users\vm\Desktop\apps\docs\Python\assignment_9_4_5.py in module()
 22 ## find the greatest number of mail messages.
 23
--- 24 maxval = none
 25 maxkee = none
 26 for kee, val in count.items():

NameError: name 'none' is not defined

In [52]: print address
c...@iupui.edu

Revised data:


## The program looks for 'From ' lines and takes the second
## word of those lines as the person who sent the mail.

fname = raw_input(Enter file name: )
handle = open (fname, 'r')
for line in handle:
if line.startswith(From: ):
address = line.split()[1]


## The program creates a Python dictionary that maps
## the sender's mail address to a count of the number
## of times they appear in the file.

count = dict()
for wrd in address:
count[wrd]= count.get(wrd,0) +1

## After the dictionary is produced, the program reads
## through the dictionary using a maximum loop to
## find the greatest number of mail messages.

maxval = none
maxkee = none
for kee, val in count.items():
if maxval == none or maxval val:
maxval = val
maxkee = kee


#items are printed

print address

Regards,
Hal


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Re: [Tutor] dictionary of lists

[Chris Stinemetz]

On Thu, Jun 4, 2015 at 2:30 AM, Peter Otten __pete...@web.de wrote:
 Chris Stinemetz wrote:

 Although I am certain it is not very efficient I was able to
 accomplish what I wanted with the following code I wrote:

 import os
 import pprint
 import csv
 from collections import defaultdict

 print_map =  {'MOU':0, 'Call_Att':1, 'Device':2}
 header = ['IMEI','MOUs','Call_Att','Device']

 path = 'C:/Users/cs062x/Desktop/Panhandle'

 os.chdir(path)
 running_MOU = {}
 call_attempts = {}
 d = defaultdict(list)
 for fname in os.listdir('.'):
 with open (fname) as csvfile:
 spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')
 next(spamreader)
 for row in spamreader:

 if row[8]:
 device = row[36]
 Elapsed_Mins = float(row[7])
 IMEI = row[8].replace(', )

 if IMEI in running_MOU.keys():

 For big dicts in Python 2 the test

 key in some_dict.keys()

 is indeed very inefficient as it builds a list of keys first and then
 performs a linear scan for the key. Much better:

 key in some_dict

 This test avoids building the list and can also use an efficient lookup
 algorithm that is independent of the size of the dict.

 running_MOU[IMEI] += Elapsed_Mins
 else:
 running_MOU[IMEI] = Elapsed_Mins

 if IMEI in call_attempts.keys():
 call_attempts[IMEI] += 1
 else:
 call_attempts[IMEI] = 1

 # if key matches append mou else append 0.
 d[IMEI] = [running_MOU[IMEI]]
 d[IMEI].append([call_attempts[IMEI]])
 d[IMEI].append([device])


 print ,.join(header)
 for k,v in sorted(d.items()):
 print k, ,, d[k][print_map['MOU']],,,
 d[k][print_map['Call_Att']][0],,, d[k][print_map['Device']][0]

 print complete

 Here's an alternative that uses only one dict:

 import csv
 import os
 import sys

 header = ['IMEI', 'MOUs', 'Call_Att', 'Device']

 path = 'C:/Users/cs062x/Desktop/Panhandle'

 d = {}
 for fname in os.listdir(path):
 with open(os.path.join(path, fname)) as csvfile:
 spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')
 next(spamreader)
 for row in spamreader:
 if row[8]:
 device = row[36]
 elapsed_mins = float(row[7])
 IMEI = row[8].replace(', )

 if IMEI in d:
 record = d[IMEI]
 record[1] += elapsed_mins
 record[2] += 1
 else:
 d[IMEI] = [IMEI, elapsed_mins, 1, device]

 writer = csv.writer(sys.stdout)
 writer.writerow(header)
 writer.writerows(sorted(d.itervalues()))

 print complete


Peter - Thank you for showing me how to do this with one dictionary
and a more efficient method to lookup dictionary keys. I originally
attempted to accomplish this by using one dictionary but could not
find a good example that is why I used the defaultdict module. Your
approach sped the parsing time up from about 3 minutes to about 15
seconds! Very cool.

Thanks,

Chris
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Re: [Tutor] dictionary of lists

[Peter Otten]

Chris Stinemetz wrote:

 Although I am certain it is not very efficient I was able to
 accomplish what I wanted with the following code I wrote:
 
 import os
 import pprint
 import csv
 from collections import defaultdict
 
 print_map =  {'MOU':0, 'Call_Att':1, 'Device':2}
 header = ['IMEI','MOUs','Call_Att','Device']
 
 path = 'C:/Users/cs062x/Desktop/Panhandle'
 
 os.chdir(path)
 running_MOU = {}
 call_attempts = {}
 d = defaultdict(list)
 for fname in os.listdir('.'):
 with open (fname) as csvfile:
 spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')
 next(spamreader)
 for row in spamreader:
 
 if row[8]:
 device = row[36]
 Elapsed_Mins = float(row[7])
 IMEI = row[8].replace(', )
 
 if IMEI in running_MOU.keys():

For big dicts in Python 2 the test 

key in some_dict.keys()

is indeed very inefficient as it builds a list of keys first and then 
performs a linear scan for the key. Much better:

key in some_dict

This test avoids building the list and can also use an efficient lookup 
algorithm that is independent of the size of the dict.

 running_MOU[IMEI] += Elapsed_Mins
 else:
 running_MOU[IMEI] = Elapsed_Mins
 
 if IMEI in call_attempts.keys():
 call_attempts[IMEI] += 1
 else:
 call_attempts[IMEI] = 1
 
 # if key matches append mou else append 0.
 d[IMEI] = [running_MOU[IMEI]]
 d[IMEI].append([call_attempts[IMEI]])
 d[IMEI].append([device])
 
 
 print ,.join(header)
 for k,v in sorted(d.items()):
 print k, ,, d[k][print_map['MOU']],,,
 d[k][print_map['Call_Att']][0],,, d[k][print_map['Device']][0]
 
 print complete

Here's an alternative that uses only one dict:

import csv
import os
import sys

header = ['IMEI', 'MOUs', 'Call_Att', 'Device']

path = 'C:/Users/cs062x/Desktop/Panhandle'

d = {}
for fname in os.listdir(path):
with open(os.path.join(path, fname)) as csvfile:
spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')
next(spamreader)
for row in spamreader:
if row[8]:
device = row[36]
elapsed_mins = float(row[7])
IMEI = row[8].replace(', )

if IMEI in d:
record = d[IMEI]
record[1] += elapsed_mins
record[2] += 1
else:
d[IMEI] = [IMEI, elapsed_mins, 1, device]

writer = csv.writer(sys.stdout)
writer.writerow(header)
writer.writerows(sorted(d.itervalues()))

print complete


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Re: [Tutor] dictionary of lists

[Alan Gauld]

On 03/06/15 17:39, Chris Stinemetz wrote:

I am trying to create a dictionary of lists as I read a file. I
envision it looking like: {key: [float_type],[string_type]}



Thats not a dictionary of lists. You maybe mean:

{key: [[float_type],[string_type]]}

Which is a dictionary of lists of lists?


For the first item in the list I am trying to add the value to the
existing value where the key matches


Sorry, I'm sure that made sense to you but not to me.
Which value are you adding to which existing value?
Can you give a before/after example?


Resetting execution engine
Running C:\Users\cs062x\Desktop\python\projects\PanHandle\PanHandle\PanHandle.py
The Python REPL process has exited


That's slightly unusual. How are you running this?


Traceback (most recent call last):
   File 
C:\Users\cs062x\Desktop\python\projects\PanHandle\PanHandle\PanHandle.py,
line 22, in module
 d[IMEI] += Elapsed_Mins
TypeError: 'float' object is not iterable





d = defaultdict(list)
for fname in os.listdir('.'):
 with open (fname) as csvfile:
 spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')
 next(spamreader)
 for row in spamreader:

 if row[8]:
 device = row[36]
 Elapsed_Mins = float(row[7])
 IMEI = row[8].replace(', )


So IMEA is a string and Elapsed_Mins is a float and d is a default dict 
that sets its defaults to lists.



 d[IMEI] += Elapsed_Mins ## this is where the error occurs.


So this is trying to add a float to a list.
 L = []
 L += f
Traceback (most recent call last):
  File stdin, line 1, in module
TypeError: 'float' object is not iterable

look familiar?

Now, what I don't know, is what you are trying to do.
Are you trying to append the float to the list?
Or to replace the list with the float?
Or to add the float to the value of the first(or last?)
element in the list - if it exists
(and if it doesn't? Then what?)


--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos


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Re: [Tutor] dictionary of lists

[Chris Stinemetz]

 Resetting execution engine
 Running
 C:\Users\cs062x\Desktop\python\projects\PanHandle\PanHandle\PanHandle.py
 The Python REPL process has exited


 That's slightly unusual. How are you running this?


I am running it with Microsoft Visual Studio Community 2013 using
Python Tools for Visual Studio



 Now, what I don't know, is what you are trying to do.
 Are you trying to append the float to the list?
 Or to replace the list with the float?
 Or to add the float to the value of the first(or last?)
 element in the list - if it exists
 (and if it doesn't? Then what?)



Although I am certain it is not very efficient I was able to
accomplish what I wanted with the following code I wrote:

import os
import pprint
import csv
from collections import defaultdict

print_map =  {'MOU':0, 'Call_Att':1, 'Device':2}
header = ['IMEI','MOUs','Call_Att','Device']

path = 'C:/Users/cs062x/Desktop/Panhandle'

os.chdir(path)
running_MOU = {}
call_attempts = {}
d = defaultdict(list)
for fname in os.listdir('.'):
with open (fname) as csvfile:
spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')
next(spamreader)
for row in spamreader:

if row[8]:
device = row[36]
Elapsed_Mins = float(row[7])
IMEI = row[8].replace(', )

if IMEI in running_MOU.keys():
running_MOU[IMEI] += Elapsed_Mins
else:
running_MOU[IMEI] = Elapsed_Mins

if IMEI in call_attempts.keys():
call_attempts[IMEI] += 1
else:
call_attempts[IMEI] = 1

# if key matches append mou else append 0.
d[IMEI] = [running_MOU[IMEI]]
d[IMEI].append([call_attempts[IMEI]])
d[IMEI].append([device])


print ,.join(header)
for k,v in sorted(d.items()):
print k, ,, d[k][print_map['MOU']],,,
d[k][print_map['Call_Att']][0],,, d[k][print_map['Device']][0]

print complete
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Re: [Tutor] dictionary keys

[Peter Otten]

Alex Kleider wrote:

 On 2014-04-08 14:34, Peter Otten wrote:
 
 That's a change in Python 3 where dict.keys() no longer creates a list,
 but
 instead creates a view on the underlying dict data thus saving time and
 space. In the rare case where you actually need a list you can
 explicitly
 create one with
 
 ips = list(ipDic)

 Thanks, Peter, for this clarification.  I want to present the list
 sorted so probably this is the rare case of which you spoke where I
 would need to use l = list(myDict) rather than the view.

You can create and sort the list in a single step:

l = sorted(myDict)

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Re: [Tutor] dictionary keys

[Alex Kleider]

On 2014-04-08 23:55, Peter Otten wrote:


You can create and sort the list in a single step:

l = sorted(myDict)



Thank you again; this is a new idiom for me.
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Re: [Tutor] dictionary keys

[Peter Otten]

Alex Kleider wrote:

 I've got a fairly large script that uses a dictionary (called 'ipDic')
 each
 value of which is a dictionary which in turn also has values which are
 not
 simple types.
 Instead of producing a simple list,
 
 ips = ipDic.keys()
 print(ips)
 
 yields
 
 dict_keys(['61.147.107.120', '76.191.204.54', '187.44.1.153'])
 
 
 Searching my code for 'dict_keys' yields nothing.  I've no idea where it
 comes from.
 
 
 
 Can anyone shed light on why instead of getting the list I'm expecting
 I
 get dict_keys( list I'm expecting )?
 
 (Using Python3, on Ubuntu  12.4)

That's a change in Python 3 where dict.keys() no longer creates a list, but 
instead creates a view on the underlying dict data thus saving time and 
space. In the rare case where you actually need a list you can explicitly 
create one with

ips = list(ipDic)

 I've been unable to reproduce this behaviour using simpler dictionaries
 which seem to work as I expect:

 d = dict(a=1, b=2, c=3)
 d
 {'a': 1, 'c': 3, 'b': 2}
 d.keys()
 ['a', 'c', 'b']
 print(d.keys())
 ['a', 'c', 'b']

That's because the above is a session using Python 2. Compare:

$ python3
Python 3.3.2+ (default, Feb 28 2014, 00:52:16) 
[GCC 4.8.1] on linux
Type help, copyright, credits or license for more information.
 dict(a=1, b=2).keys()
dict_keys(['b', 'a'])

$ python2
Python 2.7.5+ (default, Feb 27 2014, 19:37:08) 
[GCC 4.8.1] on linux2
Type help, copyright, credits or license for more information.
 dict(a=1, b=2).keys()
['a', 'b']

PS: You can get a view in Python 2, too, with dict.viewkeys()

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Re: [Tutor] dictionary keys

[Alex Kleider]

On 2014-04-08 14:34, Peter Otten wrote:

That's a change in Python 3 where dict.keys() no longer creates a list, 
but

instead creates a view on the underlying dict data thus saving time and
space. In the rare case where you actually need a list you can 
explicitly

create one with

ips = list(ipDic)




That's because the above is a session using Python 2. Compare:

$ python3
Python 3.3.2+ (default, Feb 28 2014, 00:52:16)
[GCC 4.8.1] on linux
Type help, copyright, credits or license for more information.

dict(a=1, b=2).keys()

dict_keys(['b', 'a'])

$ python2
Python 2.7.5+ (default, Feb 27 2014, 19:37:08)
[GCC 4.8.1] on linux2
Type help, copyright, credits or license for more information.

dict(a=1, b=2).keys()

['a', 'b']

PS: You can get a view in Python 2, too, with dict.viewkeys()



Thanks, Peter, for this clarification.  I want to present the list 
sorted so probably this is the rare case of which you spoke where I 
would need to use l = list(myDict) rather than the view.

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Re: [Tutor] Dictionary from a text file

[Peter Otten]

Nitish Kunder wrote:

 I have a dictionary which is in this format
 for ex:
 
 {
 '5x' : {
 '50' : {
 'update' : {
 'update-from-esxi5.0-5.0_update01' : {
 'call' : Update,
 'name' : 'Update50u1',
 'release' : '15/03/12'
 },
 'update-from-esxi5.0-5.0_update02' : {
 'call' : Update,
 'name' : 'Update50u2',
 'release' : '21/12/12'
 },
 },
  'patch' : {
 'ESXi500-201109001' : {
 'call' : Patch,
 'name' :'Patch_501',
 'release' : '13/09/11'
 },
 'ESXi500-20001' : {
 'call' : Patch,
 'name' :'ExpressPatch501',
 'release' : '13/09/11'
 },
  },
 },
  '51' : {
 'update' : {
 'update-from-esxi5.1-5.1_update01' : {
 'call' : Update,
 'name' : 'Update51u1',
 'release' : '25/04/13'
 },
 },
 'patch' : {
 'ESXi510-201210001' : {
 'call' : Patch,
 'name' :'ExpressPatch511',
 'release' : '29/08/13'
 },
 'ESXi510-201212001' : {
 'call' : Patch,
 'name' :'Patch_511',
 'release' : '20/12/12'
 },
  },
 },
 },
  }
 
 Note: in* 'call' : Update* ,Update it is a function defined in my python
 script. My dictionary is too large so i taught rather than using directly
 in python program I save it in a text file and when needed i assign it to
 dictionary object . How can i assign this text file to dictionary object
 and call it?

You could modify the dict a bit to comply with the json format and then load 
and post-process it to replace names with functions:

import json

def Update(): pass
def Patch(): pass

def post_process(x, names):
for k, v in x.items():
if k == ucall:
x[k] = names[v]
elif isinstance(v, dict):
post_process(v, names)

with open(tmp.json) as instream:
d = json.load(instream)
post_process(d, names={uPatch: Patch, uUpdate: Update})

Another option is to use a function similar to ast.literal_eval() that can 
do name lookups. Here's my adaptation of literal_eval():

import ast

def Update(): pass
def Patch(): pass

def safe_eval(s, names={}):
node = ast.parse(s, mode=eval)
if isinstance(node, ast.Expression):
node = node.body
def convert(node):
if isinstance(node, ast.Str):
return node.s
elif isinstance(node, ast.Dict):
return {convert(k): convert(v) for k, v in zip(node.keys, 
node.values)}
elif isinstance(node, ast.Name):
try:
return names[node.id]
except KeyError:
raise ValueError(Unresolved name {!r}.format(node.id))

raise ValueError(Malformed node or string {!r}.format(node))
return convert(node)

with open(tmp.txt) as instream:
data = instream.read()

d = safe_eval(data, dict(Update=Update, Patch=Patch))


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Re: [Tutor] Dictionary from a text file

[bob gailer]

On 10/31/2013 2:16 AM, Nitish Kunder wrote:

I have a dictionary which is in this format
for ex:

{
'5x' : {
'50' : {
'update' : {
'update-from-esxi5.0-5.0_update01' : {
'call' : Update,
'name' : 'Update50u1',
'release' : '15/03/12'
},
'update-from-esxi5.0-5.0_update02' : {
'call' : Update,
'name' : 'Update50u2',
'release' : '21/12/12'
},
},
'patch' : {
'ESXi500-201109001' : {
'call' : Patch,
'name' :'Patch_501',
'release' : '13/09/11'
},
'ESXi500-20001' : {
'call' : Patch,
'name' :'ExpressPatch501',
'release' : '13/09/11'
},
},
},

[snip]
Does it have to be in a dictionary format? I'd rather use sqlite to 
store the data.


Also please give us a use case. There is probably a much simper and more 
elegant solution.


--
Bob Gailer
919-636-4239
Chapel Hill NC

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Re: [Tutor] Dictionary from a text file

[Danny Yoo]

 Note: in* 'call' : Update* ,Update it is a function defined in my python
 script. My dictionary is too large so i taught rather than using directly
 in python program I save it in a text file and when needed i assign it to
 dictionary object . How can i assign this text file to dictionary object
 and call it?


To introduce some terms: you are serializing some data --- your
dictionary --- to disk and back.  There are a few libraries that help you
do the hard work of translating the nested dictionary structure to some
flat string; in addition, they almost always includes a parser to go back
from that flat string back to the nested dictionary structure.  The 'json'
library that Petter Otten suggests is one of these serialization libraries.
 Try looking into that and see if you can just take advantage of it.

Are there other consumers for this data besides just your program?
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Re: [Tutor] Dictionary get method

[Phil]

On 20/03/13 14:54, Amit Saha wrote:

Hello Phil,

On Wed, Mar 20, 2013 at 12:54 PM, Phil phil_...@bigpond.com wrote:

Thank you for reading this.

I'm working my way through a series of exercises where the author only
provides a few solutions.

The reader is asked to modify the histogram example so that it uses the get
method thereby eliminating the if and else statements. Histogram2 is my
effort.

The resulting dictionary only contains the default value provided by get
and I cannot see how the value can be incremented without an if statement.


You are almost there. Note that all you have to do is increment 1 to
the current 'value' for the key denoted by c. If you change the line
with get() to the following, it works as you want it to:

  d[c]= 1 + d.get(c, 0)

Output:

{'a': 1, 'b': 1, 'o': 2, 'n': 1, 's': 2, 'r': 2, 'u': 1, 't': 1}

histogram2
{'a': 1, 'b': 1, 'o': 2, 'n': 1, 's': 2, 'r': 2, 'u': 1, 't': 1}

You were almost there. Good Luck.

-Amit.



Thanks Amit and Mitya,

I thought I must have been close.

I've played with C++ since the mid 90s and I'm finding Python very 
refreshing.


--
Regards,
Phil
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Re: [Tutor] Dictionary get method

[Phil]

On 20/03/13 15:09, Mitya Sirenef wrote:
cut



By the way, you can further simplify it by doing:

def histogram2(s):
 return {c: d.get(c,0)+1 for c in s}


That will work in python 3, in python 2 you need:

 return dict((c: d.get(c,0)+1) for c in s)



Thanks again Mitya, although I'm not sure it's a simplification at my 
present level.


--
Regards,
Phil
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Re: [Tutor] Dictionary get method

[Peter Otten]

Phil wrote:

 On 20/03/13 15:09, Mitya Sirenef wrote:
 cut
 

 By the way, you can further simplify it by doing:

 def histogram2(s):
  return {c: d.get(c,0)+1 for c in s}


 That will work in python 3, in python 2 you need:

  return dict((c: d.get(c,0)+1) for c in s)

 
 Thanks again Mitya, although I'm not sure it's a simplification at my
 present level.

Especially as Mitya's code doesn't work.

 {k: v for k, v in [(1, a), (2, b)]}
{1: 'a', 2: 'b'}

is called dict comprehension, it builds a dict from a list of key-value 
pairs. However, there is no way to reference the resulting dict while it is 
being built, and that is necessary for a histogram. It is possible to use 
dict.update() with a generator expression

 d = {}
 d.update((c, d.get(c, 0)+1) for c in abba)
 d
{'a': 2, 'b': 2}

but frankly, I don't see how that is better than the for loop.

So as your experience with Python grows you may continue to use a loop or 
switch to the standard library's collections.Counter (Python3 only).

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Re: [Tutor] Dictionary get method

[Mitya Sirenef]

On 03/20/2013 04:21 AM, Peter Otten wrote:

Phil wrote:


On 20/03/13 15:09, Mitya Sirenef wrote:
cut


By the way, you can further simplify it by doing:

def histogram2(s):
  return {c: d.get(c,0)+1 for c in s}


That will work in python 3, in python 2 you need:

  return dict((c: d.get(c,0)+1) for c in s)


Thanks again Mitya, although I'm not sure it's a simplification at my
present level.

Especially as Mitya's code doesn't work.



Ah, yes - I messed up here.. I agree the loop is the best option here,
vs. the example below. -m


d = {}
d.update((c, d.get(c, 0)+1) for c in abba)
d

{'a': 2, 'b': 2}

but frankly, I don't see how that is better than the for loop.

So as your experience with Python grows you may continue to use a loop or
switch to the standard library's collections.Counter (Python3 only).

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--
Lark's Tongue Guide to Python: http://lightbird.net/larks/

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Re: [Tutor] Dictionary get method

[Mitya Sirenef]

On 03/19/2013 10:54 PM, Phil wrote:

Thank you for reading this.


 I'm working my way through a series of exercises where the author 
only provides a few solutions.


 The reader is asked to modify the histogram example so that it uses 
the get method thereby eliminating the if and else statements. 
Histogram2 is my effort.


 The resulting dictionary only contains the default value provided by 
get and I cannot see how the value can be incremented without an if 
statement.


 def histogram(s):
 d = dict()
 for c in s:
 if c not in d:
 d[c] = 1
 else:
 d[c] += 1
 return d

 def histogram2(s):
 d = dict()
 for c in s:
 d[c]= d.get(c, 0)

 return d

 h = histogram(brontosaurs)

 print h

 print

 print histogram2

 h = histogram2(brontosaurs)

 print h



Add + 1

 -m


--
Lark's Tongue Guide to Python: http://lightbird.net/larks/

Whenever you find yourself on the side of the majority, it is time to
pause and reflect.  Mark Twain

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Re: [Tutor] Dictionary get method

[Amit Saha]

Hello Phil,

On Wed, Mar 20, 2013 at 12:54 PM, Phil phil_...@bigpond.com wrote:
 Thank you for reading this.

 I'm working my way through a series of exercises where the author only
 provides a few solutions.

 The reader is asked to modify the histogram example so that it uses the get
 method thereby eliminating the if and else statements. Histogram2 is my
 effort.

 The resulting dictionary only contains the default value provided by get
 and I cannot see how the value can be incremented without an if statement.

You are almost there. Note that all you have to do is increment 1 to
the current 'value' for the key denoted by c. If you change the line
with get() to the following, it works as you want it to:

 d[c]= 1 + d.get(c, 0)

Output:

{'a': 1, 'b': 1, 'o': 2, 'n': 1, 's': 2, 'r': 2, 'u': 1, 't': 1}

histogram2
{'a': 1, 'b': 1, 'o': 2, 'n': 1, 's': 2, 'r': 2, 'u': 1, 't': 1}

You were almost there. Good Luck.

-Amit.

-- 
http://amitsaha.github.com/
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Re: [Tutor] Dictionary get method

[Mitya Sirenef]

On 03/19/2013 10:54 PM, Phil wrote:

Thank you for reading this.


 I'm working my way through a series of exercises where the author 
only provides a few solutions.


 The reader is asked to modify the histogram example so that it uses 
the get method thereby eliminating the if and else statements. 
Histogram2 is my effort.


 The resulting dictionary only contains the default value provided by 
get and I cannot see how the value can be incremented without an if 
statement.


 def histogram(s):
 d = dict()
 for c in s:
 if c not in d:
 d[c] = 1
 else:
 d[c] += 1
 return d

 def histogram2(s):
 d = dict()
 for c in s:
 d[c]= d.get(c, 0)

 return d

 h = histogram(brontosaurs)

 print h

 print

 print histogram2

 h = histogram2(brontosaurs)

 print h


By the way, you can further simplify it by doing:

def histogram2(s):
return {c: d.get(c,0)+1 for c in s}


That will work in python 3, in python 2 you need:

return dict((c: d.get(c,0)+1) for c in s)


--
Lark's Tongue Guide to Python: http://lightbird.net/larks/

Those who can make you believe absurdities can make you commit
atrocities.  Voltaire


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Re: [Tutor] Dictionary get method

[Mitya Sirenef]

On 03/20/2013 01:09 AM, Mitya Sirenef wrote:



By the way, you can further simplify it by doing:

def histogram2(s):
return {c: d.get(c,0)+1 for c in s}


That will work in python 3, in python 2 you need:

return dict((c: d.get(c,0)+1) for c in s)



Sorry, it should use a comma:

return dict((c, d.get(c,0)+1) for c in s)

-m


--
Lark's Tongue Guide to Python: http://lightbird.net/larks/

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Re: [Tutor] Dictionary

[bob gailer]

On 6/17/2012 2:26 PM, Selby Rowley-Cannon wrote:
[snip]
Do you have any programming (algorithm development) experience?

Do you want to translate words independent of context?

--
Bob Gailer
919-636-4239
Chapel Hill NC

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Re: [Tutor] Dictionary

[James Reynolds]

Does this language have grammar independent of english?

If no, just use .split() on the string and loop through that.

If yes, well, its much more complicated
On Jun 17, 2012 2:27 PM, Selby Rowley-Cannon selbyrowleycan...@ymail.com
wrote:

 Version: 2.7
 OS: Ubuntu 12.04 LTS

 I am writing a small translation app for Rydish (A language that exists in
 the same way Klingon does, invented by my company for a[n] RPG).
 Here is my current method of translation:

 Edictionary = {'English keys':'Rydish values'}
 TextEng = raw_input('Please enter your text: ')
 if TextEng in Edictionary:
 print(TextEng + ' traslates to ' + Edictionary[TextEng])

 But I have found that this is only going to translate one word at a time.
 I thought about a loop of somesort, but I can't seem to find one that won't
 still force the user to translate one word at a time. Can anyone tell me
 how to translate a full sentance/several sentances?

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Re: [Tutor] dictionary of methods calling syntax

[Gregory, Matthew]

Alan Gauld wrote:
 Since a class is effectively a disguised dictionary I'm not sure why you
 want to do this? If you just want to access the method by name then why
 not just call getattr(spam,'get_mean')?

Thanks for the feedback and, yes, this makes sense.  My use case was when the 
statistic desired was going to be specified at runtime (through file input or 
UI) and that a dictionary would be a convenient crosswalk to associate the 
statistic name with the method name (and thus avoid an if/else ladder).  But I 
understand that as long as there is a direct relationship between the name of 
the statistic and my class method (e.g. 'mean' - get_mean), that I should be 
able to use the getattr() syntax as above.

Thanks also to Joel for the suggestion to put the dictionary inside of __init__.

thanks, matt
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Re: [Tutor] dictionary of methods calling syntax

[Mark Lawrence]

On 08/02/2012 17:41, Gregory, Matthew wrote:

Alan Gauld wrote:

Since a class is effectively a disguised dictionary I'm not sure why you
want to do this? If you just want to access the method by name then why
not just call getattr(spam,'get_mean')?


Thanks for the feedback and, yes, this makes sense.  My use case was when the 
statistic desired was going to be specified at runtime (through file input or UI) 
and that a dictionary would be a convenient crosswalk to associate the statistic 
name with the method name (and thus avoid an if/else ladder).  But I understand 
that as long as there is a direct relationship between the name of the statistic 
and my class method (e.g. 'mean' -  get_mean), that I should be able to use 
the getattr() syntax as above.

Thanks also to Joel for the suggestion to put the dictionary inside of __init__.

thanks, matt
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This should help if you need more info 
http://code.activestate.com/lists/python-list/403361/


--
Cheers.

Mark Lawrence.

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Re: [Tutor] dictionary of methods calling syntax

[Joel Goldstick]

On Tue, Feb 7, 2012 at 2:32 PM, Gregory, Matthew
matt.greg...@oregonstate.edu wrote:
 Hi list,

 I'm trying to understand how to use a class-level dictionary to act as a 
 switch for class methods.  In the contrived example below, I have the 
 statistic name as the key and the class method as the value.

 class Statistics(object):
    STAT = {
        'MEAN': get_mean,
        'SUM': get_sum,
    }
    def __init__(self, a, b):
        self.a = a
        self.b = b
    def get_mean(self):
        return (self.a + self.b) / 2.0
    def get_sum(self):
        return (self.a + self.b)
    def get_stat(self, stat):
        f = self.STAT[stat.upper()]
        return f(self)

 if __name__ == '__main__':
    spam = Statistics(4, 3)
    print spam.get_stat('mean')
    print spam.get_stat('sum')

 When I try to run this, I get:

  NameError: name 'get_mean' is not defined

 If I move the STAT dictionary to the bottom of the class, it works fine.  I 
 understand why I get an error, i.e. when the dictionary is created get_mean 
 hasn't yet been defined, but I'm wondering if there is a better common 
 practice for doing this type of lookup.  My web searches didn't come up with 
 anything too applicable.

 thanks, matt
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You might want to read through this:
http://stackoverflow.com/questions/5213166/python-forward-declaration-of-functions-inside-classes

Specifically from an answer:
First, in class B, the function foo() is called before being
declared. A does not have this problem because foo() is only called
when the class is instantiated--after the function foo is defined.

So, I think you could move your STAT dictionary definition into the
__init__ method so that is doesn't actually run until you create an
instance of your class.  That being said, I'm not sure that is more
'pythonic' than moving it to the bottom of the class definition

-- 
Joel Goldstick
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Re: [Tutor] dictionary of methods calling syntax

[Alan Gauld]

On 07/02/12 19:32, Gregory, Matthew wrote:


class Statistics(object):
 STAT = {
 'MEAN': get_mean,
 'SUM': get_sum,
 }

...


if __name__ == '__main__':
 spam = Statistics(4, 3)
 print spam.get_stat('mean')
 print spam.get_stat('sum')



Since a class is effectively a disguised dictionary I'm not sure why you 
want to do this? If you just want to access the method by name then why 
not just call getattr(spam,'get_mean')?


--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/

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Re: [Tutor] Dictionary to variable copy

[Timo]

Op 08-12-11 10:03, sunil tech schreef:
/Can i copy the content if a dictionary in to another variable, with 
out any link between the dictionary  the variable?

/

Have a look at the copy module [1], or use the builtin dict.copy() method.

Cheers,
Timo

[1] http://docs.python.org/library/copy.html#module-copy


/
if so, please teach.
/


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Re: [Tutor] Dictionary to variable copy

[Alan Gauld]

On 08/12/11 09:03, sunil tech wrote:

/Can i copy the content if a dictionary in to another variable, with out
any link between the dictionary  the variable?

if so, please teach.


Yes, but they will both refer to the same object.
But the two references will be entirely independant:

D = {1:2,3:4}
d = D[1]   # both d and D[1] reference 2

D[1] = 42  # d still refers to 2

del(D) # d still refers to 2

Is that what you mean?

Or do you wantbto make a clone of the value so that you
can change it without changing the original?

In that case you can use several techniques depending on object type. 
Generically use copy or deepcopy.


HTH
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/

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Re: [Tutor] Dictionary File character reader

[Noah Hall]

On Tue, May 10, 2011 at 5:27 AM, Clara Mintz jamani.s...@hotmail.com wrote:
 Sorry I am completely new at python and don't understand why this function
 is returning an empty dictionary. I want it to take a list of files open
 them then return the number of characters as the value and the file name as
 the key.
 def fileLengths(files):
     d = {}
     files = []

Here's why. You take the name files and assign it to an empty list,
so you lose the files that you sent to the function.
 files = [file_1.txt, file_2.doc]
 files
['file_1.txt', 'file_2.doc']
 files = []
 files
[]
You don't need that line at all, so remove it, and see what happens. :)

     for file in files:
         reader = open(file)
         for line in reader:
             char = sum(len(line))

As a side not, this doesn't need to be sum in this case - the len is
simply len, throwing away what it previously was. In fact, this
probably isn't what you want. At any one point, char is the length of
a single line, not the sum of every length of every line in the file.
What you want is something that takes the length of each line, and add
it to the sum. A simple way would be to do
sum(len(line) for line in file)

         d[file] = char
         reader.close
     return d
 Thank you sorry I don't understand what I am doing wrong.
 -Clara
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Hope this helped.
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Re: [Tutor] Dictionary File character reader

[Alan Gauld]

Noah Hall enali...@gmail.com wrote 

What you want is something that takes the length of each line, 
and add it to the sum. A simple way would be to do

sum(len(line) for line in file)


And if you just want the total count for the file an even 
simpler way is to use file.read()


count = len(file.read())


One other thing to consider is whether you want to include 
line feed characters in the count...


HTH,


--
Alan Gauld
Author of the Learn to Program web site
http://www.alan-g.me.uk/


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Re: [Tutor] Dictionary File character reader

[Dave Angel]

On 01/-10/-28163 02:59 PM, Clara Mintz wrote:


Sorry I am completely new at python and don't understand why this function is 
returning an empty dictionary. I want it to take a list of files open them then 
return the number of characters as the value and the file name as the key.
def fileLengths(files):d = {}files = []for file in files:
reader = open(file)for line in reader:char = sum(len(line)) 
   d[file] = charreader.closereturn d
Thank you sorry I don't understand what I am doing wrong.
-Clara  


The first thing you're doing is wordwrapping your program fragment.  It 
makes the code hard to read, and some aspects impossible, as we can't 
tell what parts were indented by how much.


The second thing is clobbering your input parameter.  files=[] will 
obliterate whatever argument was passed to that function.


You will have more problems after that:
   sum() won't work on an integer, so it's not clear why you're calling 
len() on the line.
  You're returning d from the function, but nothing in the function 
ever inserts anything into it.  So it will clearly be empty.


print is your friend.  A couple of carefully placed print statements 
would reveal these problems.


DaveA
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Re: [Tutor] dictionary

[naheed arafat]

Supposing your dictionary like this: dict={1:'My name is X',2:'My name is x
y z',3: 'i am X'}
You can use len(list) :
 dict={1:'My name is X',2:'My name is x y z',3: 'i am X'}
 for values in dict.values():
...  if len(values.split(' '))3:
...print values
My name is X
My name is x y z

On Fri, May 6, 2011 at 7:56 AM, louis leichtnam l.leicht...@gmail.comwrote:

 HEllo everyone,

 I have a dictionnary, and I would like to print only the values that have
 more/equal than 3 spaces in them for example: My name is Xavier.

 Can you help me out,

 Thanks

 Louis

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Re: [Tutor] dictionary

[Steven D'Aprano]

louis leichtnam wrote:

HEllo everyone,

I have a dictionnary, and I would like to print only the values that have
more/equal than 3 spaces in them for example: My name is Xavier.



d = {1: Hello world, 2: My name is Xavier, 3: ham and eggs,
 4: Eat more cheese please!, 5: spam spam spam spam spam spam,
 6: how much wood would a woodchuck chuck if a woodchuck would
  chuck wood?, 7: nice cup of tea and a slice of cake}
for value in d.values():
if value.count( ) = 3:
print(value)



--
Steven

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Re: [Tutor] Dictionary Question

[Steven D'Aprano]

Garry Bettle wrote:

Howdy all,

Hope this message finds everyone well.

I have dictionary of keys and a string of values.

i.e.

8 Fixtures:


I assume each fixture is a key, e.g. Swin, HGrn, etc.


Swin1828 1844 1901 1916 1932 1948 2004 2019 2036 2052 2107 2122
HGrn1148 1204 1218 1232 1247 1304 1319 1333 1351
Newc1142 1157 1212 1227 1242 1258 1312 1327 1344 1403
Yarm1833 1849 1906 1922 1937 1953 2009 2024 2041 2057 2112
BVue1418 1437 1457 1517 1538 1558 1618 1637 1657 1717 1733 1747 1804 181
Hove1408 1427 1447 1507 1528 1548 1608 1627 1647 1707 1722 1738 1756 181
Romfd   1930 1946 2003 2019 2035 2053 2109 2125 2141 2157 2213 2230
Sund1839 1856 1911 1927 1943 1958 2014 2031 2047 2102 2117

I print that with the following:

f = open(SummaryFile, 'a')
header = %d Fixtures, %d Races:\n % len(FixtureDict.keys())
print header

f.write(header)
f.write(\n)
for fixture, racetimes in FixtureDict.iteritems():
line = %s\t%s  % (fixture,  .join(racetimes))



According to your description, racetimes is already a single string, so 
using join on it would be the wrong thing to do:


 racetimes = 1839 1856 1911
  .join(racetimes)
'1 8 3 9   1 8 5 6   1 9 1 1'


So what is racetimes? Is it a string, or is it a list of strings?

['1839', '1856', '1911']

I'm going to assume the latter. That's the right way to do it.



print line
f.write(line + \n)
f.write(\n)
f.close()

What I'd like to is add the number of values to the Header line.

So how would I get i.e.

8 Fixtures, 93 Races

I tried

header = %d Fixtures, %d Races:\n % (len(FixtureDict.keys()),
len(FixtureDict.values()))

But I get

print header

8 Fixture, 8 Races


Any ideas?


You need len(racetimes) rather than len(FixtureDict.values()).

Every dict has exactly one value for every key, always, without 
exception. That is, len(dict.keys()) == len(dict.values()). In this 
case, the values are lists of multiple start times, but each list counts 
as one value. You need to count the number of items inside each value, 
not the number of values.


In this case, you need to sum the number of races for all the fixtures:

num_races = sum(len(racetimes) for racetimes in FixtureDict.values())




--
Steven

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Re: [Tutor] Dictionary Question

[bob gailer]

On 12/22/2010 7:31 AM, Steven D'Aprano wrote:

Also note: len(dict.keys()) == len(dict.values()) == len(dict)

--
Bob Gailer
919-636-4239
Chapel Hill NC

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Re: [Tutor] Dictionary Question

[Garry Bettle]

On Wed, 22 Dec 2010 23:31:39 +1100, Steven D'Aprano wrote:
 In this case, you need to sum the number of races for all the fixtures:

 num_races = sum(len(racetimes) for racetimes in FixtureDict.values())

Many thanks Steven for your explanation and final golden nugget of code.

On Wed, 22 Dec 2010 10:11:25 -0500, bob gailer wrote:

 Also note: len(dict.keys()) == len(dict.values()) == len(dict)

Yup, thanks Bob.

Cheers,

G.
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Re: [Tutor] dictionary and more

[Jack Uretsky]

Hi-
	I apologize for the size of the attached file, but the question 
would make little sense if I made substantial deletions.

The program gives the following error:
Traceback (most recent call last):
  File /Javastuff/python/Glau_Is.py, line 131, in module
Adestin[state]()
TypeError: 'NoneType' object is not callable
	This is after the RETA prinout from function A_1, so that seems to 
be called correctly.  The black screen is not part of the problem, I am 
not including the .jpeg files.  Line 131 is the one that reads

Adestin[state]()
and the program prints (correctly) state=1 before the traceback.
Thanks in advance for any insights.
JLU

Trust me.  I have a lot of experience at this.
General Custer's unremembered message to his men,
just before leading them into the Little Big Horn Valley



 
import pygame
from pygame.locals import *
from sys import exit
import time
import random
from math import *
global gamma
gamma = .25
global RET
RET = [0,0]
global state
global r
def A_1():
state =1
r = 1 - gamma
f = random.randrange(3)
g = f+2  #destination
RET = (g,r)
print RETA!=, RET
def A_2():
state = 2
global r2
r2 = 1. + gamma
f = random.randrange(3)
if f == 0:
g = 1 #destination
elif f == 1:
g,r2 = 7,1.
elif f == 2:
g,r2 = 6,1
else:
print alarm 2
RET = [g,r2]
def A_3():
state =3
global r3
r3 = 1. + gamma
f = random.randrange(3)
if f == 1:
g = 1
elif f == 0:
g,r3 = 7, 1.
elif f == 2:
g, r3 = 5, 1
else:
print alarm 3
RET = [g, r3]
def A_4():
state = 4
global r4
r4 = 1. + gamma
f = random.randrange(3)
if f == 2:
g = 1
elif f == 0:
g,r4 = 6,1
elif f == 1:
g,r4 = 5,1
else:
print alarm 4 
RET = [g, r4]
def A_5():
state =5
global r5
r5 = 1. + gamma
f = random.randrange(3)
if f == 0:
g = 8
elif f == 1:
g,r5 = 4,1
elif f == 2:
g,r5 = 3,1
else:
print alarm 5
RET = [g, r5]
def A_6():
state = 6
global r6
r6 = 1. + gamma
f= random.randrange(3)
if f == 1:
g = 8
elif f == 0:
g,r6 = 4,1
elif f == 2:
g,r6 = 3,1
else:
print alarm 6
RET = [g, r6]
def A_7():
state = 7
global r7
r7 = 1. + gamma
f = random.randrange(3)
if f == 2:
g = 8
elif f == 0:
g,r7 = 3,1
elif f == 1:
g,r7 = 2,1
else:
print alarm 7
RET = [g, r7]
def A_8():
state = 8
global r8
r8 = 1. - gamma
f = random.randrange(3)
g = f + 5
RET = [g, r8]
#dictionaries
Adestin = {1: A_1(), 2 : A_2(), 3 : A_3(), 4: A_4(),5 : A_5(), 6 : A_6(), 7: 
A_7(), 8 :
 A_8()}
apics = {1: 'a_1.jpg', 2: 'a_2.jpg', 3 : 'a_3.jpg', 4: 'a_4.jpg',
 5: 'a_5.jpg', 6: 'a_6.jpg', 7 : 'a_7.jpg', 8 : 'a_8.jpg'}
state = 1 #set initial state
pygame.init()
screen = pygame.display.set_mode((640,480), 0, 32)
pygame.event.pump()
back1 =apics[state]
back = pygame.image.load(back1).convert()
screen.blit(back, (0,0))
pygame.display.update()
RET=(1,1-gamma)
k=0
while k = 5:
k = k+1
back1 = apics[state]
print state=, state
Adestin[state]()   
print RET, k
destin = RET[0]
pois = RET[1]
back = pygame.image.load(back1).convert()
screen.blit(back, (0,0))
pygame.display.update()
state = destin
print state=,destin
time.sleep(5)
print end
   
 
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Re: [Tutor] dictionary and more

[Alan Gauld]

Jack Uretsky j...@hep.anl.gov wrote 

Please do not reply to an existing thread when starting a new topic. 
This messes up threaded newsreaders and several archives. 
Start a new topic with a newmessage.


 I apologize for the size of the attached file, but the question 
would make little sense if I made substantial deletions.

 The program gives the following error:
Traceback (most recent call last):
  File /Javastuff/python/Glau_Is.py, line 131, in module
Adestin[state]()
TypeError: 'NoneType' object is not callable


This says that Adestin[state] is None. 
So lets see how Adestin gets initialised

Its a function call, which returns None. So Python is correct.
You want to store the function names (more accurately the 
function objects) so you do not want the () after the name 
when you initialise the dictionary.


Here is an example:

def f1(): print 'f1'

def f2(): print 'f2'

dct = {'1': f1, '2': f2}  # notice no parens, only a reference

for key in dct:
   dct[key]() # use the parens here to call the func


HTH,


--
Alan Gauld
Author of the Learn to Program web site
http://www.alan-g.me.uk/


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Re: [Tutor] dictionary and more

[Emile van Sebille]

On 10/11/2010 9:19 AM Jack Uretsky said...

Hi-
I apologize for the size of the attached file, but the question would
make little sense if I made substantial deletions.
The program gives the following error:
Traceback (most recent call last):
File /Javastuff/python/Glau_Is.py, line 131, in module
Adestin[state]()
TypeError: 'NoneType' object is not callable
This is after the RETA prinout from function A_1, so that seems to be
called correctly. The black screen is not part of the problem, I am not
including the .jpeg files. Line 131 is the one that reads
Adestin[state]()
and the program prints (correctly) state=1 before the traceback.


Then Adestin[1] is None causing TypeError: 'NoneType' object is not 
callable.  You can verify that by printing Adestin[1] where you're 
currently printing state.


Then you'd discover that when you're building Adestin you're assigning 
the _results_ of the functions to your dictionary.  You probably want to 
leave off the parens when declaring the functions.  Further, your 
functions probably need to return something -- add return statements if 
you want the results.


HTH,

Emile

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Re: [Tutor] dictionary and more

[Steven D'Aprano]

On Tue, 12 Oct 2010 03:19:22 am Jack Uretsky wrote:
 Hi-
   I apologize for the size of the attached file, but the question
 would make little sense if I made substantial deletions.
   The program gives the following error:
 Traceback (most recent call last):
File /Javastuff/python/Glau_Is.py, line 131, in module
  Adestin[state]()
 TypeError: 'NoneType' object is not callable

The question makes perfect sense without even looking at the attached 
file. Adestin[state] returns None, which you attempt to call as a 
function. The questions you should be asking yourself are:

- what value am I expecting it to contain?
- where in my code should that value be set?
- why is it setting None instead?

Armed with these three questions, you now know how to debug the problem 
yourself.



-- 
Steven D'Aprano
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Re: [Tutor] Dictionary Comprehensions

[Christian Witts]

Khalid Al-Ghamdi wrote:

Hi everyone!

I'm using python 3.1 and I want to to know why is it when I enter the 
following in a dictionary comprehension:


 dc={y:x for y in list(khalid) for x in range(6)}

I get the following:
{'a': 5, 'd': 5, 'i': 5, 'h': 5, 'k': 5, 'l': 5}

instead of the expected:
{'a': 0, 'd': 1, 'i': 2, 'h': 3, 'k': 4, 'l': 5}

and is there a way to get the target (expected) dictionary using a 
dictionary comprehension. 


note that I tried sorted(range(6)) also but to no avail.

thanks 




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Are you maybe looking for `dc = {y:x for y,x in zip('khalid', range(6))}` ?

--
Kind Regards,
Christian Witts


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Re: [Tutor] Dictionary Comprehensions

[spir]

Lie Ryan lie.1...@gmail.com dixit:

 note that:
   [(y, x) for y in list(khalid) for x in range(6)]  
 [('k', 0), ('k', 1), ('k', 2), ('k', 3), ('k', 4), ('k', 5), ('h', 0), 
 ('h', 1), ('h', 2), ('h', 3), ('h', 4), ('h', 5), ('a', 0), ('a', 1), 
 ('a', 2), ('a', 3), ('a', 4), ('a', 5), ('l', 0), ('l', 1), ('l', 2), 
 ('l', 3), ('l', 4), ('l', 5), ('i', 0), ('i', 1), ('i', 2), ('i', 3), 
 ('i', 4), ('i', 5), ('d', 0), ('d', 1), ('d', 2), ('d', 3), ('d', 4), 
 ('d', 5)]
 
 and when that big list is turned into a dict it gives:
   dict(_)  
 {'a': 5, 'd': 5, 'i': 5, 'h': 5, 'k': 5, 'l': 5}

... because a dict holds a single value per key, so last value overrides 
previous ones.

Denis


la vita e estrany

http://spir.wikidot.com/
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Re: [Tutor] Dictionary Comprehensions

[spir]

Hugo Arts hugo.yo...@gmail.com dixit:

 bc = {y: x for x, y in enumerate(khalid)}
 
 Note that your output is like so:
 {'a': 2, 'd': 5, 'i': 4, 'h': 1, 'k': 0, 'l': 3}
 
 The first character in your original string gets a zero, the second a
 one, so on and so forth. I'm hoping that's what you meant. If you
 really want this:
 
 {'a': 0, 'd': 1, 'i': 2, 'h': 3, 'k': 4, 'l': 5}
 
 I'm not sure how to do that programmatically. 

# first need a list of sorted chars
# otherwise python cannot guess what order you mean:
chars = sorted(list(khalid))
print chars# == ['a', 'd', 'h', 'i', 'k', 'l']

# enumerate gives a list of (index, value) pairs
# from which you can construct a dict:
#~ dc = {index:char for (index,char) in enumerate(chars)}
# or (python version  3)
dc = dict((index,char) for (index,char) in enumerate(chars))
print dc   # == {0: 'a', 1: 'd', 2: 'h', 3: 'i', 4: 'k', 5: 'l'}

Denis


la vita e estrany

http://spir.wikidot.com/
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Re: [Tutor] Dictionary Comprehensions

[Lie Ryan]

On 12/5/2009 7:32 AM, Khalid Al-Ghamdi wrote:

Hi everyone!

I'm using python 3.1 and I want to to know why is it when I enter the
following in a dictionary comprehension:

  dc={y:x for y in list(khalid) for x in range(6)}


are you sure you want this?
{'a': 0, 'd': 1, 'i': 2, 'h': 3, 'k': 4, 'l': 5}

instead of:
{'a': 2, 'd': 5, 'i': 4, 'h': 1, 'k': 0, 'l': 3}

for the former case, you can't, you can't guarantee any sort of ordering 
in dictionary. You should use ordered dictionary instead.


For the latter case, it's easy with zip()
dc={y:x for x, y in zip(khalid, range(6))}

as for why python did that, it's because dictionary comprehension is 
supposed to have similar semantic with:

dc = {x: y for x, y in lst}
dc = dict(  (x, y) for x, y in lst  )

so this:
dc={y:x for y in list(khalid) for x in range(6)}
becomes:
dc=dict(  (y, x) for y in list(khalid) for x in range(6)  )

note that:
 [(y, x) for y in list(khalid) for x in range(6)]
[('k', 0), ('k', 1), ('k', 2), ('k', 3), ('k', 4), ('k', 5), ('h', 0), 
('h', 1), ('h', 2), ('h', 3), ('h', 4), ('h', 5), ('a', 0), ('a', 1), 
('a', 2), ('a', 3), ('a', 4), ('a', 5), ('l', 0), ('l', 1), ('l', 2), 
('l', 3), ('l', 4), ('l', 5), ('i', 0), ('i', 1), ('i', 2), ('i', 3), 
('i', 4), ('i', 5), ('d', 0), ('d', 1), ('d', 2), ('d', 3), ('d', 4), 
('d', 5)]


and when that big list is turned into a dict it gives:
 dict(_)
{'a': 5, 'd': 5, 'i': 5, 'h': 5, 'k': 5, 'l': 5}

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Re: [Tutor] Dictionary Comprehensions

[Hugo Arts]

On Fri, Dec 4, 2009 at 9:32 PM, Khalid Al-Ghamdi emailkg...@gmail.com wrote:
 Hi everyone!
 I'm using python 3.1 and I want to to know why is it when I enter the
 following in a dictionary comprehension:
 dc={y:x for y in list(khalid) for x in range(6)}
 I get the following:
 {'a': 5, 'd': 5, 'i': 5, 'h': 5, 'k': 5, 'l': 5}
 instead of the expected:
 {'a': 0, 'd': 1, 'i': 2, 'h': 3, 'k': 4, 'l': 5}
 and is there a way to get the target (expected) dictionary using a
 dictionary comprehension.
 note that I tried sorted(range(6)) also but to no avail.
 thanks

That dictionary comprehension is equivalent to the following code:

dc = {}
for x in range(6):
for y in list(khalid):
dc[y] = x

This makes it clear what is wrong. The two for loops come out as
nested, rather than zipped.
The general fix for something like this is the zip function:

bc = {x: y for x, y in zip(khalid, xrange(6))}

However, in this case, the idiomatic way to write this would be the
enumerate function:

bc = {y: x for x, y in enumerate(khalid)}

Note that your output is like so:
{'a': 2, 'd': 5, 'i': 4, 'h': 1, 'k': 0, 'l': 3}

The first character in your original string gets a zero, the second a
one, so on and so forth. I'm hoping that's what you meant. If you
really want this:

{'a': 0, 'd': 1, 'i': 2, 'h': 3, 'k': 4, 'l': 5}

I'm not sure how to do that programmatically. The dict object prints
its objects in no particular order, so figuring out that order is hard
(and very likely implementation/platform dependent). My best guess was
sorted(khalid, key=hash):

{'a': 0, 'd': 1, 'i': 3, 'h': 2, 'k': 4, 'l': 5}

close, but no cigar. Anyone who can think of a clever hack for this?
Not that it's very useful, but fun.

Hugo
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Re: [Tutor] Dictionary Comprehensions

[Emile van Sebille]

On 12/4/2009 12:32 PM Khalid Al-Ghamdi said...

Hi everyone!

I'm using python 3.1 and I want to to know why is it when I enter the 
following in a dictionary comprehension:


  dc={y:x for y in list(khalid) for x in range(6)}


Try breaking this into pieces...

First see what [(x,y) for y in in list(khalid) for x in range(6)]
gets you, then see how that fits into dict().

To get where you want, take a look at zip'ing the two lists.

(II don't remember -- s zip still in 3.1?)

Emile




I get the following:
{'a': 5, 'd': 5, 'i': 5, 'h': 5, 'k': 5, 'l': 5}

instead of the expected:
{'a': 0, 'd': 1, 'i': 2, 'h': 3, 'k': 4, 'l': 5}

and is there a way to get the target (expected) dictionary using a 
dictionary comprehension. 


note that I tried sorted(range(6)) also but to no avail.

thanks 





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Re: [Tutor] Dictionary Comprehensions

[Dave Angel]

Khalid Al-Ghamdi wrote:

Hi everyone!

I'm using python 3.1 and I want to to know why is it when I enter the
following in a dictionary comprehension:

  

dc={y:x for y in list(khalid) for x in range(6)}



I get the following:
{'a': 5, 'd': 5, 'i': 5, 'h': 5, 'k': 5, 'l': 5}

instead of the expected:
{'a': 0, 'd': 1, 'i': 2, 'h': 3, 'k': 4, 'l': 5}

and is there a way to get the target (expected) dictionary using a
dictionary comprehension.

note that I tried sorted(range(6)) also but to no avail.

thanks

  
You're confused about what two for loops do here.  It's basically a 
doubly-nested loop, with the outer loop iterating from k through d, and 
the inner loop iterating from 0 to 5.  So there are 36 entries in the 
dictionary, but of course the dictionary overwrites all the ones with 
the same key.  For each letter in the outer loop, it iterates through 
all six integers, and settles on 5.


To do what you presumably want, instead of a doubly nested loop you need 
a single loop with a two-tuple for each item, consisting of one letter 
and one digit.


  dc = { y:x for y,x in zip(khalid, range(6)) }

The output for this is:
 {'a': 2, 'd': 5, 'i': 4, 'h': 1, 'k': 0, 'l': 3}

Now, this isn't the same values for each letter as you expected, but 
I'm not sure how you came up with that particular order.I expect, and 
get, 0 for the first letter 'k' and 1 for the 'h'.  etc.


Perhaps printing out the zip would make it clearer:
   list( zip(khalid, range(6)) )
yields
   [('k', 0), ('h', 1), ('a', 2), ('l', 3), ('i', 4), ('d', 5)]

DaveA

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Re: [Tutor] Dictionary, integer, compression

[Alan Gauld]

Dinesh B Vadhia dineshbvad...@hotmail.com wrote

Say, you have a dictionary of integers, are the integers stored 
in a compressed integer format or as integers ie. are integers 
encoded before being stored in the dictionary and then 
decoded when read?


I can't think of any reason to compress them, I imagine they 
are stored as integers. But given the way Python handlers 
integers with arbitrarily long numbers etc it may well be more 
complex than a simple integer (ie 4 byte number). But any 
form of compression would be likely to hit performamce 
so I doubt that they would be compressed.


Is there anything that made you think they might be?

HTH


--
Alan Gauld
Author of the Learn to Program web site
http://www.alan-g.me.uk/

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Re: [Tutor] Dictionary, integer, compression

[Stefan Behnel]

Dinesh B Vadhia wrote:
 Say, you have a dictionary of integers, are the integers stored in a
 compressed integer format or as integers ie. are integers encoded before
 being stored in the dictionary and then decoded when read?

Integer objects are not special cased in dictionaries. They are stored as
normal int/long objects.

Dictionaries do not use any kind of compression in Python.

Stefan

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Re: [Tutor] Dictionary, integer, compression

[Dinesh B Vadhia]

Alan

I want to perform test runs on my local machine with very large numbers of 
integers stored in a dictionary.  As the Python dictionary is an built-in 
function I thought that for very large dictionaries there could be compression. 
 Done correctly, integer compression wouldn't affect performance but could 
enhance it.  Weird, I know!  I'll check in with the comp.lang.python lot.

Dinesh




Message: 3
Date: Wed, 29 Apr 2009 17:35:53 +0100
From: Alan Gauld alan.ga...@btinternet.com
Subject: Re: [Tutor] Dictionary, integer, compression
To: tutor@python.org
Message-ID: gt9vl7$oh...@ger.gmane.org
Content-Type: text/plain; format=flowed; charset=iso-8859-1;
reply-type=original


Dinesh B Vadhia dineshbvad...@hotmail.com wrote

 Say, you have a dictionary of integers, are the integers stored 
 in a compressed integer format or as integers ie. are integers 
 encoded before being stored in the dictionary and then 
 decoded when read?

I can't think of any reason to compress them, I imagine they 
are stored as integers. But given the way Python handlers 
integers with arbitrarily long numbers etc it may well be more 
complex than a simple integer (ie 4 byte number). But any 
form of compression would be likely to hit performamce 
so I doubt that they would be compressed.

Is there anything that made you think they might be?

HTH


-- 
Alan Gauld
Author of the Learn to Program web site
http://www.alan-g.me.uk/

 
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Re: [Tutor] dictionary looping problem

[Steve Willoughby]

On Tue, Dec 02, 2008 at 01:08:09PM -0800, Jeremiah Jester wrote:
 Hello,
 
 I'm trying to gather a list of files and md5 hash them to do a checksum.
 I've created a function for each dictionary. However, when i print out
 the dictionary I don't get all the items. Any ideas?

Yep.  Don't use os.system() there.  

1. you're running the md5 program externally when you don't need to,
   since Python has the ability to compute md5 checksums on its own,
   which you already know because you imported that module at the top
   of your script (and then didn't use).

2. The return value from os.system() is NOT the hash, so what you're
   storing in the dictionary is not going to be that unique, and so
   each call which yields the same return value (0, usually) will 
   overwrite that element in the dict.

-- 
Steve Willoughby|  Using billion-dollar satellites
[EMAIL PROTECTED]   |  to hunt for Tupperware.
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