Word frequency statistics (string, list)

Count the number of occurrences of each letter in the string
1. utilize Counter function

from collections import Counter
c = Counter("loonbalxballpoon")
print(c)

2. Building an empty dictionary , Traverse

str1="loonbalxballpoon"
gg={
    }
for i in str1:
    gg[i]=0
for i in str1:
    gg[i]+=1
print(gg)

result ：

Counter({
    'l': 4, 'o': 4, 'n': 2, 'b': 2, 'a': 2, 'x': 1, 'p': 1})

leetcode242：
Given two strings s and t , Write a function to determine t Whether it is s Letter heteronym of .
Be careful ： if s and t Each character in the has the same number of occurrences , said s and t They are mutually alphabetic words .
Example 1:
Input : s = “anagram”, t = “nagaram”
Output : true

def isAnagram(s: str, t: str) -> bool:
        letter_frequcy_ransomNote = {
    }
        for i in range(len(s)):
            letter_frequcy_ransomNote[s[i]] = 0
        for i in range(len(s)):
            letter_frequcy_ransomNote[s[i]] += 1
        letter_frequcy_magazine = {
    }
        for i in range(len(t)):
            letter_frequcy_magazine[t[i]] = 0
        for i in range(len(t)):
            letter_frequcy_magazine[t[i]] += 1
        if letter_frequcy_magazine==letter_frequcy_ransomNote:
            return True
        else:
            return False

leetcode1189：
Give you a string text, You need to use text To piece together as many words as possible “balloon”（ balloon ）.
character string text Each letter in can only be used once at most . Please return the maximum number of words you can piece together “balloon”.

def maxNumberOfBalloons(text: str) -> int:
        # Method 1 ：
        # dict_text = {'b':0,'a':0,'l':0,'o':0,'n':0}
        # for x in text:
        # if x in dict_text:
        # dict_text[x] += 1
        # dict_text['l'] //= 2
        # dict_text['o'] //= 2
        # return min(dict_text.values())
        # Method 2 ：
        c = Counter(text)
        c['l']//=2
        c['o']//=2
        return min(c['l'],c['o'],c['b'],c['n'],c['a'])

3. utilize jieba

list1=jieba.lcut(" I love Tsinghua University and Peking University ")
#jieba Of lcut Function to segment a string into words and phrases , It's a list back 
print(list1)
print(Counter(list1))

result ：

[' I ', ' Love ', ' Tsinghua University ', ' and ', ' Peking University, ']
Counter({
    ' I ': 1, ' Love ': 1, ' Tsinghua University ': 1, ' and ': 1, ' Peking University, ': 1})

Count the number of times each letter appears in the list
The same way of thinking

d=Counter(['aa','bb','cc','aa'])
print(d)

Counter({
    'aa': 2, 'bb': 1, 'cc': 1})