通过放入字典来计算列表中的值

Question

alist = [5, 7, 6, 2, 9, 1, 7]

D = {}

for each unique number in list, set a new key for each key in dictionary, count number of that key and set to value

this should look like {5:1, 2:1, 6:1, 9:1, 1:1, 7:2}

algorithm:

For each number n on the input list:
∗ If n in count: set count[n] to count[n] + 1
∗ else: set count[n] to 1

I don't know how to go about this. Can anybody show me how?

Attempt:

for number in alist:
    if number in D:
        D[number] = D[number]+1
    else:
        D[number] = 1

Error:

Traceback (most recent call last): File "<pyshell#15>", line 3, in
<module> D[number] = D[number]+1 KeyError: 5

Answer 1

>>> from collections import Counter
>>> Counter(alist)
Counter({7: 2, 1: 1, 2: 1, 5: 1, 6: 1, 9: 1})

Answer 2

Simplest way to do this is a dictionary comprehension

count_dict = {elem:a_list.count(elem) for elem in a_list}

It's like a for loop but it specifies both the key and value for each dictionary item.

Answer 3

Ayush Shanker's answer is a perfectly valid way to do what you want. I'll point out how to fix the code you have now.

First, in these two lines

for number in alist:
    if number in alist:

You're basically saying "for every number in alist, check to see if it's in alist". Of course it is! So, you really want if number in D: for the second line, because then you're checking to see if the number is in the dictionary you're building.

The second problem is this:

        D[number] = D[number+1]

number and number+1 here are indices, so you're referring to two different elements in D . The solution is to move the +1 outside. This is what the fixed code looks like:

for number in alist:
    if number in D:
        D[number] = D[number] + 1
    else:
        D[number] = 1

To be more Pythonic, you can also replace D[number] = D[number] + 1 with D[number] += 1 .

Answer 4

To fix you code and algorithm, you need to check if n is in D.keys() (which returns list of keys of D dictionary:

>>> D = {}
>>> for n in alist:
    if n in D.keys():
        D[n] += 1
    else:
        D[n] = 1


>>> D
{1: 1, 2: 1, 5: 1, 6: 1, 7: 2, 9: 1}

Answer 5

SOLUTION:

 def mode(somelist):
        count = {}
        for n in alist:
            if n in count:
                count[n]+=1
            else:
                count[n] = 1
        return(count)