Re: [Tutor] Rearranging a list of numbers with corresponding index (RESOLVED)
On 03/13/2015 11:55 AM, Ken G. wrote: I will be studying this also. I am still using Python 2.7.6 as that is the latest as provided by Ubuntu 14.04.1. FYI, Python 3.4 is installed by default in Ubuntu 14.04(.1). You have to invoke it as 'python3' though. MMR... ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Rearranging a list of numbers with corresponding index
On 03/13/2015 09:57 AM, Ken G. wrote: I have been keeping track of numbers drawn in our local lotto drawings into a list format as shown in a short example below. Using such list, I am able to determine how often a number appears within the last 100 plus drawings. The length of my lists range from 5, 15, 35, 59and 75 long. I will give an example of one of my short list. PowerPlay = [0, 0, 61, 32, 11, 14] Disregarding index 0 and 1, I can see and print the following: IndexNumber 2 61 3 32 4 11 5 14 showing that number 2 appears 61 times, number 3 appears 32 times, number 4 appears 11 times and number 5 appears 14 times within last 100 plus drawings. How I best rearrange the numbers from high to low with its corresponding index number such as below: Number Index 612 323 14 5 11 4 showing the number 2 appears 61 times, number 3 appears 32 times, number 5 appears 14 times and number 4 appears 11 times. I know that using MegaBall.reverse() sort the number from high to low but the index numbers still remain in the same positions. Thanks for pointing out the way to do this. Make a list of tuples by doing something like: new_list = [ (cnt, index) for index, cnt in enumerate(power_play) ] to produce: [ (0, 0), (0, 1), (61, 2), (32, 3), (11, 4), (14, 5) ] then sort it with reverse=True. When you print it, just omit the items that have zero as their count. (they'll be at the end, anyway) Lots of other variants, some more concise. And other approaches might be cleaner if we change how we generate the list. (I tried it briefly using Python 3.4) -- DaveA ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Rearranging a list of numbers with corresponding index
On 03/13/2015 10:21 AM, Peter Otten wrote: Ken G. wrote: I have been keeping track of numbers drawn in our local lotto drawings into a list format as shown in a short example below. Using such list, I am able to determine how often a number appears within the last 100 plus drawings. The length of my lists range from 5, 15, 35, 59and 75 long. I will give an example of one of my short list. PowerPlay = [0, 0, 61, 32, 11, 14] Disregarding index 0 and 1, I can see and print the following: IndexNumber 2 61 3 32 4 11 5 14 showing that number 2 appears 61 times, number 3 appears 32 times, number 4 appears 11 times and number 5 appears 14 times within last 100 plus drawings. How I best rearrange the numbers from high to low with its corresponding index number such as below: Number Index 612 323 14 5 11 4 showing the number 2 appears 61 times, number 3 appears 32 times, number 5 appears 14 times and number 4 appears 11 times. I know that using MegaBall.reverse() sort the number from high to low but the index numbers still remain in the same positions. Thanks for pointing out the way to do this. The initial list: power_play = [0, 0, 61, 32, 11, 14] Use enumerate() to get index, frequency pairs: list(enumerate(power_play)) [(0, 0), (1, 0), (2, 61), (3, 32), (4, 11), (5, 14)] Sort the pairs: sorted(enumerate(power_play)) [(0, 0), (1, 0), (2, 61), (3, 32), (4, 11), (5, 14)] Provide a key to sort after the second value, i. e. the frequency in the (index, frequency) pairs: def second_value(item): ... return item[1] ... sorted(enumerate(power_play), key=second_value) [(0, 0), (1, 0), (4, 11), (5, 14), (3, 32), (2, 61)] Note that instead of writing your own custom function or lambda you could also use operator.itemgetter(1) from the operator module in the stdlib. Sort in reverse order: sorted(enumerate(power_play), key=second_value, reverse=True) [(2, 61), (3, 32), (5, 14), (4, 11), (0, 0), (1, 0)] Print the output for frequencies 0: pairs = sorted(enumerate(power_play), key=second_value, reverse=True) for index, freq in pairs: ... if freq == 0: ... break ... print(freq, index, sep=\t) ... 61 2 32 3 14 5 11 4 Wow! Thanks! Printing this out and will be studying it completely. Again, thanks. Ken ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
[Tutor] Rearranging a list of numbers with corresponding index
I have been keeping track of numbers drawn in our local lotto drawings into a list format as shown in a short example below. Using such list, I am able to determine how often a number appears within the last 100 plus drawings. The length of my lists range from 5, 15, 35, 59and 75 long. I will give an example of one of my short list. PowerPlay = [0, 0, 61, 32, 11, 14] Disregarding index 0 and 1, I can see and print the following: IndexNumber 2 61 3 32 4 11 5 14 showing that number 2 appears 61 times, number 3 appears 32 times, number 4 appears 11 times and number 5 appears 14 times within last 100 plus drawings. How I best rearrange the numbers from high to low with its corresponding index number such as below: Number Index 612 323 14 5 11 4 showing the number 2 appears 61 times, number 3 appears 32 times, number 5 appears 14 times and number 4 appears 11 times. I know that using MegaBall.reverse() sort the number from high to low but the index numbers still remain in the same positions. Thanks for pointing out the way to do this. Ken ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Rearranging a list of numbers with corresponding index
Ken G. wrote: I have been keeping track of numbers drawn in our local lotto drawings into a list format as shown in a short example below. Using such list, I am able to determine how often a number appears within the last 100 plus drawings. The length of my lists range from 5, 15, 35, 59and 75 long. I will give an example of one of my short list. PowerPlay = [0, 0, 61, 32, 11, 14] Disregarding index 0 and 1, I can see and print the following: IndexNumber 2 61 3 32 4 11 5 14 showing that number 2 appears 61 times, number 3 appears 32 times, number 4 appears 11 times and number 5 appears 14 times within last 100 plus drawings. How I best rearrange the numbers from high to low with its corresponding index number such as below: Number Index 612 323 14 5 11 4 showing the number 2 appears 61 times, number 3 appears 32 times, number 5 appears 14 times and number 4 appears 11 times. I know that using MegaBall.reverse() sort the number from high to low but the index numbers still remain in the same positions. Thanks for pointing out the way to do this. The initial list: power_play = [0, 0, 61, 32, 11, 14] Use enumerate() to get index, frequency pairs: list(enumerate(power_play)) [(0, 0), (1, 0), (2, 61), (3, 32), (4, 11), (5, 14)] Sort the pairs: sorted(enumerate(power_play)) [(0, 0), (1, 0), (2, 61), (3, 32), (4, 11), (5, 14)] Provide a key to sort after the second value, i. e. the frequency in the (index, frequency) pairs: def second_value(item): ... return item[1] ... sorted(enumerate(power_play), key=second_value) [(0, 0), (1, 0), (4, 11), (5, 14), (3, 32), (2, 61)] Note that instead of writing your own custom function or lambda you could also use operator.itemgetter(1) from the operator module in the stdlib. Sort in reverse order: sorted(enumerate(power_play), key=second_value, reverse=True) [(2, 61), (3, 32), (5, 14), (4, 11), (0, 0), (1, 0)] Print the output for frequencies 0: pairs = sorted(enumerate(power_play), key=second_value, reverse=True) for index, freq in pairs: ... if freq == 0: ... break ... print(freq, index, sep=\t) ... 61 2 32 3 14 5 11 4 ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Rearranging a list of numbers with corresponding index (RESOLVED)
On 03/13/2015 10:38 AM, Dave Angel wrote: On 03/13/2015 09:57 AM, Ken G. wrote: I have been keeping track of numbers drawn in our local lotto drawings into a list format as shown in a short example below. Using such list, I am able to determine how often a number appears within the last 100 plus drawings. The length of my lists range from 5, 15, 35, 59and 75 long. I will give an example of one of my short list. PowerPlay = [0, 0, 61, 32, 11, 14] Disregarding index 0 and 1, I can see and print the following: IndexNumber 2 61 3 32 4 11 5 14 showing that number 2 appears 61 times, number 3 appears 32 times, number 4 appears 11 times and number 5 appears 14 times within last 100 plus drawings. How I best rearrange the numbers from high to low with its corresponding index number such as below: Number Index 612 323 14 5 11 4 showing the number 2 appears 61 times, number 3 appears 32 times, number 5 appears 14 times and number 4 appears 11 times. I know that using MegaBall.reverse() sort the number from high to low but the index numbers still remain in the same positions. Thanks for pointing out the way to do this. Make a list of tuples by doing something like: new_list = [ (cnt, index) for index, cnt in enumerate(power_play) ] to produce: [ (0, 0), (0, 1), (61, 2), (32, 3), (11, 4), (14, 5) ] then sort it with reverse=True. When you print it, just omit the items that have zero as their count. (they'll be at the end, anyway) Lots of other variants, some more concise. And other approaches might be cleaner if we change how we generate the list. (I tried it briefly using Python 3.4) I will be studying this also. I am still using Python 2.7.6 as that is the latest as provided by Ubuntu 14.04.1. With thanks to all, I was able to resolve my request for assistance here in this forum. Thanks Peter. Ken ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Rearranging a list
I may have mis-stated my intention. I will rewrite my request for assistance later and resubmit. Thanks, Ken On 02/26/2015 08:04 AM, Peter Otten wrote: Steven D'Aprano wrote: Ah wait, the penny drops! Now I understand what you mean! Glad I'm not the only one ;) ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Rearranging a list
Steven D'Aprano wrote: Ah wait, the penny drops! Now I understand what you mean! Glad I'm not the only one ;) ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Rearranging a list
Ken G. wrote: Assuming I have the following list and code how do I best be able rearrange the list as stated below: list = [0, 0, 21, 35, 19, 42] Using print list[2:6] resulted in the following: 221 335 419 542 I would like to rearrange the list as follow: 542 335 221 419 I tried using list.reverse() and print list[0,6] and it resulted in: [42, 19, 35, 21, 0, 0] or as printed: 042 119 235 321 40 50 In another words, I would desire to show that: 5 appears 42 times 3 appears 35 times 2 appears 21 times 4 appears 19 times but then I lose my original index of the numbers by reversing. How do I best keep the original index number to the rearrange numbers within a list? Don't rearrange the original list, make a new one instead: frequencies = [0, 0, 21, 35, 19, 42] wanted_indices = [5, 3, 2, 4] [frequencies[i] for i in wanted_indices] [42, 35, 21, 19] Or look up the values as you print them: for i in wanted_indices: ... print(i, appears, frequencies[i], times) ... 5 appears 42 times 3 appears 35 times 2 appears 21 times 4 appears 19 times The expression [frequencies[i] for i in wanted_indices] is called list comprehension and is a shortcut for a loop: result = [] for i in wanted_indices: ... result.append(frequencies[i]) ... result [42, 35, 21, 19] ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Rearranging a list
Ah wait, the penny drops! Now I understand what you mean!
On Thu, Feb 26, 2015 at 07:13:57AM -0500, Ken G. wrote:
In another words, I would desire to show that:
5 appears 42 times
3 appears 35 times
2 appears 21 times
4 appears 19 times
but then I lose my original index of the numbers by reversing. How do I
best keep the original index number to the rearrange numbers within a list?
Here's your list again:
[0, 0, 21, 35, 19, 42]
I think using a list is the wrong answer. I think you should use a
dictionary:
d = {0: 0, 1: 0, 2: 21, 3: 35, 4: 19, 5: 42}
for key, value in sorted(d.items(), reverse=True):
print(key, value)
which gives this output:
5 42
4 19
3 35
2 21
1 0
0 0
How should you build the dict in the first place? There's an easy way,
and an even easier way. Here's the easy way.
# count the numbers
numbers = [2, 3, 1, 4, 0, 4, 2, 5, 2, 3, 4, 1, 1, 1, 3, 5]
counts = {}
for n in numbers:
i = counts.get(n, 0)
counts[n] = i + 1
print(counts)
which gives output:
{0: 1, 1: 4, 2: 3, 3: 3, 4: 3, 5: 2}
Here's the even easier way:
from collections import Counter
counts = Counter(numbers)
print(counts)
which gives very similar output:
Counter({1: 4, 2: 3, 3: 3, 4: 3, 5: 2, 0: 1})
In both cases, you then run the earlier code to print the items in
order:
for key, value in sorted(counts.items(), reverse=True):
print(key, value)
--
Steve
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Re: [Tutor] Rearranging a list
Peter Otten wrote: Ken G. wrote: Assuming I have the following list and code how do I best be able rearrange the list as stated below: list = [0, 0, 21, 35, 19, 42] Using print list[2:6] resulted in the following: 221 335 419 542 I would like to rearrange the list as follow: 542 335 221 419 I tried using list.reverse() and print list[0,6] and it resulted in: [42, 19, 35, 21, 0, 0] or as printed: 042 119 235 321 40 50 In another words, I would desire to show that: 5 appears 42 times 3 appears 35 times 2 appears 21 times 4 appears 19 times but then I lose my original index of the numbers by reversing. How do I best keep the original index number to the rearrange numbers within a list? Don't rearrange the original list, make a new one instead: frequencies = [0, 0, 21, 35, 19, 42] wanted_indices = [5, 3, 2, 4] [frequencies[i] for i in wanted_indices] [42, 35, 21, 19] Or look up the values as you print them: for i in wanted_indices: ... print(i, appears, frequencies[i], times) ... 5 appears 42 times 3 appears 35 times 2 appears 21 times 4 appears 19 times The expression [frequencies[i] for i in wanted_indices] is called list comprehension and is a shortcut for a loop: result = [] for i in wanted_indices: ... result.append(frequencies[i]) ... result [42, 35, 21, 19] Oops, it occurs to me that you may want the most frequent indices rather than specific indices. You can achieve that with def lookup_frequency(index): ... return frequencies[index] ... wanted_indices = sorted(range(len(frequencies)), key=lookup_frequency, reverse=True) wanted_indices [5, 3, 2, 4, 0, 1] wanted_indices[:4] [5, 3, 2, 4] You may also want to look at the collections.Counter class: import collections c = collections.Counter() for i, freq in enumerate(frequencies): ... c[i] = freq ... c.most_common(4) [(5, 42), (3, 35), (2, 21), (4, 19)] for key, freq in c.most_common(4): ... print(key, appears, freq, times) ... 5 appears 42 times 3 appears 35 times 2 appears 21 times 4 appears 19 times ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Rearranging a list
On 26/02/2015 16:33, Ken G. wrote: I may have mis-stated my intention. I will rewrite my request for assistance later and resubmit. Thanks, Ken When you come back please don't top post, it makes following long threads difficult if not impossible, thanks. -- My fellow Pythonistas, ask not what our language can do for you, ask what you can do for our language. Mark Lawrence ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
[Tutor] Rearranging a list
Assuming I have the following list and code how do I best be able rearrange the list as stated below: list = [0, 0, 21, 35, 19, 42] Using print list[2:6] resulted in the following: 221 335 419 542 I would like to rearrange the list as follow: 542 335 221 419 I tried using list.reverse() and print list[0,6] and it resulted in: [42, 19, 35, 21, 0, 0] or as printed: 042 119 235 321 40 50 In another words, I would desire to show that: 5 appears 42 times 3 appears 35 times 2 appears 21 times 4 appears 19 times but then I lose my original index of the numbers by reversing. How do I best keep the original index number to the rearrange numbers within a list? Thanking in advance, Ken ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
