[Tutor] Rearranging a list

[Steven D'Aprano]

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