Python 统计字母出现频率
# coding=utf-8from collections import Counters = "absdflsjflkdsjflkfjwelkrklewrmlkewmfslkdfjsdljflsdflkewnjklhflkjdsfdsf"print dict(Counter(s))# {'a': 1, 'b': 1, 'e': 4, 'd': 7, 'f': 11, 'h': 1, 'k': 9, 'j': 7, 'm': 2, 'l': 12, 'n': 1, 's': 8, 'r': 2, 'w': 4}如果解决了您的问题请采纳!
如果未解决请继续追问
class Solution(object): def lengthOfLongestSubstring(self, s): """ :type s: str :rtype: int """ lenS = len(s); maxLen = 0; left = right = 0; charDict = {}; while right < lenS: if s[right] in charDict and left <= charDict[s[right]]: #出现重复的字符且这个字符在窗口中 left = charDict[s[right]] + 1; #左窗口位置调整到这个重复字符的右端 else: #没有重复字符出现 maxLen = max(maxLen,right - left + 1); #更新最长长度 charDict[s[right]] = right; #更新字符位置 right += 1; #右窗口后移 return maxLen;
代码如下:
#coding=utf-8
# 输入
s = input()
# 统计
d = {}
for c in s:
d[c] = (d[c] + 1) if c in d else 1
# 输出
for i in d:
print(i, d[i])
运行结果:
排序版本:
#coding=utf-8
# 输入
s = input()
# 统计
d = {}
for c in s:
d[c] = (d[c] + 1) if c in d else 1
# 排序
result = sorted(d.items(), key = lambda x:x[1], reverse = True)
# 输出
for i in result:
print(i[0], i[1])
运行结果;
你好,你可以考虑使用下面的代码:
inputStr=“adhdbxhsjjsbbsxjdjjebbxbsheuhdbbd”
ResultDict={}
for index in range(len(inputStr)):
if inputStr[index] in ResultDict:
ResultDict[inputStr[index]] +=1
else:
ResultDict[inputStr[index]]=1
for each in ResultDict:
print(each +” “ +str(ResultDict[each]))
智旺小儿:[答案] # coding=utf-8 from collections import Counter s = "absdflsjflkdsjflkfjwelkrklewrmlkewmfslkdfjsdljflsdflkewnjklhflkjdsfdsf" print dict(Counter(s)) # {'a': 1, 'b': 1, 'e': 4, 'd': 7, 'f': 11, 'h': 1, 'k': 9, 'j': 7, 'm': 2, 'l': 12, 'n': 1, 's': 8, 'r': 2, 'w': 4}如果解决了您的问题请采...
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