使用Python计算列表中的字符串数

Question

I have a list in my hand and I want to create vocabulary from this list. Then, I want to show each word and count the same strings in this list.

The sample list as below.

    new_list = ['one', 'thus', 'once', 'one', 'count', 'once', 'this', 'thus']

First, I created a vocabulary with below.

    vocabulary = []
        for i in range (0, len(new_list)):
            if new_list[i] not in vocabulary:
                vocabulary.append(new_list[i])`
    print vocabulary

The output of above code is: "count, once, one, this, thus."

I want to show the number of each words in the list as below. [count][1], [once][2], [one][2], [this][1], [thus][2].

In order to get above result; I try below code.

    matris = []

    for i in range(0,len(new_list)):
        temp = []
        temp.insert(0,new_list.count(new_list[i]))        
        matris.append(temp)

    for x in matris:
        print x

Above code only gives the number of words. Can someone advise me how can I print the word name and number of the words together such as in [once][2] format.

Answer 1

Use a Counter dict to get the word count then just iterate over the .items :

from collections import Counter

new_list = ['one', 'thus', 'once', 'one', 'count', 'once', 'this', 'thus']

cn = Counter(new_list)
for k,v in cn.items():
    print("{} appears  {} time(s)".format(k,v))

If you want that particular output you can wrap the elements in the str.format:

for k,v in cn.items():
    print("[{}][{}]".format(k,v))

[thus][2]
[count][1]
[one][2]
[once][2]
[this][1]

To get the output from highest count to lowest use .most_common:

cn = Counter(new_list)
for k,v in cn.most_common():
    print("[{}][{}]".format(k,v))

Output:

[once][2]
[thus][2]
[one][2]
[count][1]
[this][1]

If you want the data alphabetically from lowest to highest and from highest to lowest for the count you need to pass a key -x[1] to sorted to negate the count sorting the count from highest to lowest:

for k, v in sorted(cn.items(), key=lambda x: (-x[1],x[0])):
    print("[{}][{}]".format(k, v))

Output:

[once][2]
[one][2]
[thus][2]
[count][1]
[this][1]