[Tutor] finding digit in string

2012-10-08 Thread Benjamin Fishbein 
I figured out a solution for the question I asked on here.
To find the next digit (0-9) in a string, I use:
text.find(0 or 1 or 2, etc...)
But is there a more elegant way to do this? The way I found uses a lot of 
typing.
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Re: [Tutor] finding digit in string

2012-10-08 Thread Mark Lawrence 

On 08/10/2012 17:43, Benjamin Fishbein wrote:

I figured out a solution for the question I asked on here.
To find the next digit (0-9) in a string, I use:
text.find(0 or 1 or 2, etc...)
But is there a more elegant way to do this? The way I found uses a lot of 
typing.
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How about something like:-

for ch in text:
if ch.isdigit():
doSomething(ch)

or:-

for ch in text:
if '0' = ch = '9':
doSomething(ch)


If you want to use the offset for any reason use enumerate:-

for i, ch in enumerate(text):
etc.

--
Cheers.

Mark Lawrence.

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Re: [Tutor] finding digit in string

2012-10-08 Thread Alan Gauld 

On 08/10/12 17:43, Benjamin Fishbein wrote:

I figured out a solution for the question I asked on here.
To find the next digit (0-9) in a string, I use:
text.find(0 or 1 or 2, etc...)


Are you sure that worked? It doesn't for me on Python 2.7...

 s
'a string with 7 words in it'
 s.find('4' or '5' or '7')
-1
 s.find('7')
14


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

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Re: [Tutor] finding digit in string

2012-10-08 Thread Dave Angel 
On 10/08/2012 12:43 PM, Benjamin Fishbein wrote:
 I figured out a solution for the question I asked on here.

Why then did you start a new thread, instead of responding on the same
one?  You didn't even use the same subject string.

 To find the next digit (0-9) in a string, I use:
 text.find(0 or 1 or 2, etc...)

That won't work.  0 or 1  is just 0 So you're searching for
the character zero.

 But is there a more elegant way to do this? The way I found uses a lot of 
 typing.


See the other thread you started, with subject
[Tutor] finding a number with str.find

-- 

DaveA

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Re: [Tutor] finding digit in string

2012-10-08 Thread bob gailer 

On 10/8/2012 12:43 PM, Benjamin Fishbein wrote:

I figured out a solution for the question I asked on here.
To find the next digit (0-9) in a string, I use:
text.find(0 or 1 or 2, etc...)
 But is there a more elegant way to do this? The way I found uses a 
lot of typing.


My way is:

import string
tt = string.maketrans('0123456789','00')
ts = asdf3456'.translate(t) # yields 'asdf'
ts.find(0) # finds the next 0 which is in the same position as 
corresponding digit


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

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Re: [Tutor] finding digit in string

2012-10-08 Thread Prasad, Ramit 
Mark Lawrence wrote:
 On 08/10/2012 17:43, Benjamin Fishbein wrote:
  I figured out a solution for the question I asked on here.
  To find the next digit (0-9) in a string, I use:
  text.find(0 or 1 or 2, etc...)
  But is there a more elegant way to do this? The way I found uses a lot of 
  typing.
 
 
 How about something like:-
 
 for ch in text:
  if ch.isdigit():
  doSomething(ch)
 
 or:-
 
 for ch in text:
  if '0' = ch = '9':
  doSomething(ch)

I am not sure that will work very well with Unicode numbers. I would 
assume (you know what they say about assuming) that str.isdigit() 
works better with international characters/numbers.

 '1'.isdigit()
True
 '23'.isdigit()
True
 '23a'.isdigit()
False

for ch in text:
if ch.isdigit():
# do something

 
 
 If you want to use the offset for any reason use enumerate:-
 
 for i, ch in enumerate(text):
  etc.
 
 --
 Cheers.
 
 Mark Lawrence.

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Re: [Tutor] finding digit in string

2012-10-08 Thread eryksun 
On Mon, Oct 8, 2012 at 4:11 PM, Prasad, Ramit ramit.pra...@jpmorgan.com wrote:

 for ch in text:
  if '0' = ch = '9':
  doSomething(ch)

 I am not sure that will work very well with Unicode numbers. I would
 assume (you know what they say about assuming) that str.isdigit()
 works better with international characters/numbers.

In my tests below, isdigit() matches both decimal digits ('Nd') and
other digits ('No'). None of the 'No' category digits works with
int().

Python 2.7.3

 chars = [unichr(i) for i in xrange(sys.maxunicode + 1)]
 digits = [c for c in chars if c.isdigit()]
 digits_d = [d for d in digits if category(d) == 'Nd']
 digits_o = [d for d in digits if category(d) == 'No']
 len(digits), len(digits_d), len(digits_o)
(529, 411, 118)

Decimal

 nums = [int(d) for d in digits_d]
 [nums.count(i) for i in range(10)]
[41, 42, 41, 41, 41, 41, 41, 41, 41, 41]

Other

 print u''.join(digits_o[:3] + digits_o[12:56])
²³¹⁰⁴⁵⁶⁷⁸⁹₀₁₂₃₄₅₆₇₈₉①②③④⑤⑥⑦⑧⑨⑴⑵⑶⑷⑸⑹⑺⑻⑼⒈⒉⒊⒋⒌⒍⒎⒏⒐
 print u''.join(digits_o[67:94])
❶❷❸❹❺❻❼❽❾➀➁➂➃➄➅➆➇➈➊➋➌➍➎➏➐➑➒
 print u''.join(digits_o[3:12])
፩፪፫፬፭፮፯፰፱

 int(digits_o[67])
Traceback (most recent call last):
  File stdin, line 1, in module
UnicodeEncodeError: 'decimal' codec can't encode character
u'\u2776' in position 0: invalid decimal Unicode string


Python 3.2.3

 chars = [chr(i) for i in range(sys.maxunicode + 1)]
 digits = [c for c in chars if c.isdigit()]
 digits_d = [d for d in digits if category(d) == 'Nd']
 digits_o = [d for d in digits if category(d) == 'No']
 len(digits), len(digits_d), len(digits_o)
(548, 420, 128)

Decimal

 nums = [int(d) for d in digits_d]
 [nums.count(i) for i in range(10)]
[42, 42, 42, 42, 42, 42, 42, 42, 42, 42]

Other

 print(*(digits_o[:3] + digits_o[13:57]), sep='')
²³¹⁰⁴⁵⁶⁷⁸⁹₀₁₂₃₄₅₆₇₈₉①②③④⑤⑥⑦⑧⑨⑴⑵⑶⑷⑸⑹⑺⑻⑼⒈⒉⒊⒋⒌⒍⒎⒏⒐
 print(*digits_o[68:95], sep='')
❶❷❸❹❺❻❼❽❾➀➁➂➃➄➅➆➇➈➊➋➌➍➎➏➐➑➒
 print(*digits_o[3:12], sep='')
፩፪፫፬፭፮፯፰፱

 int(digits_o[68])
Traceback (most recent call last):
  File stdin, line 1, in module
ValueError: invalid literal for int() with base 10: '❶'
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