How to create an array of 10 digits for counting repetitions?

Question

I wondered how to create an array of 10 digits, such that each time a digit of a number shows, it increments its matching place in the 10 digits array, like this:

digits=[0,0,0,0,0,0,0,0,0,0]
num_digits=[1,2,3,9,1]

and digits becomes:

digits=[0,2,1,1,0,0,0,0,0,1]

I tried:

digits[num_digits[j]]=digits[num_digits[j]]+1

(j goes backwards on num_digits elements) but I got the error "list indices must be integers, not list".

Thanks in advance!

Answer 1

>>> digits=[0,0,0,0,0,0,0,0,0,0]
>>> num_digits=[1,2,3,9,1]
>>> for d in num_digits:
...     digits[d] += 1
... 
>>> digits
[0, 2, 1, 1, 0, 0, 0, 0, 0, 1]

We wouldn't need a variable j for a simple iteration of a list.

Answer 2

It might be worth using a dict rather than a list to store your totals. You could then use defaultdict from collections to create a dict which will automatically create a entry with a zero int for new keys:

>>> from collections import defaultdict
>>> digits = defaultdict(int)
>>> num_digits = [1,2,3,9,1]
>>> for d in num_digits:
...     digits[d] += 1
... 
>>> digits
defaultdict(<type 'int'>, {1: 2, 2: 1, 3: 1, 9: 1})
>>> digits[1]
2
>>> digits[8]
0

Answer 3

You can use collections.Counter for that task as follows:

from collections import Counter

num_digits=[1,2,3,9,1]
digits = [0]*10
for key, value in Counter(num_digits).items():
    digits[key] = value

Answer 4

在这里，您可以找到如何在python中实现计数排序： http : //en.wikibooks.org/wiki/Algorithm_Implementation/Sorting/Counting_sort#Python