Python 字典，设置并计数检查频率是否可以相同
原文:https://www . geesforgeks . org/python-dictionary-set-counter-check-frequency-can-变成/
给定一个包含较低字母字符的字符串，我们需要从该字符串中最多删除一个字符，这样字符串中每个不同字符的频率就变得相同。
示例:
Input : str = “xyyz”
Output : Yes
We can remove character ’y’ from above
string to make the frequency of each
character same.
Input : str = “xyyzz”
Output : Yes
We can remove character ‘x’ from above
string to make the frequency of each
character same.
Input : str = “xxxxyyzz”
Output : No
It is not possible to make frequency of
each character same just by removing at
most one character from above string.
此问题已有解决方案，请参考检查一次移除链接是否可以使所有字符的频率变得相同。我们将在 Python 中快速解决这个问题。方法很简单，
- 我们需要统计字符串中每个字母的频率，为此我们将使用 Counter(input) 方法，它返回一个以字符为键，以它们各自的频率为值的字典。
- 现在提取每个字符的频率列表，并在 python 中的 Set() 数据结构中推送这些值。
-
Since set contains unique values, so if size of set is 1 that means frequencies of all characters were same, if size of set is 2 then check if value of first element is 1 or not ( if 1 then we can make same frequency by removing one character at most otherwise not possible ).
```py
Function to Check if frequency of all characters
can become same by one removal
from collections import Counter
def allSame(input):
# calculate frequency of each character # and convert string into dictionary dict=Counter(input)
# now get list of all values and push it # in set same = list(set(dict.values()))
if len(same)>2: print('No') elif len (same)==2 and same[1]-same[0]>1: print('No') else: print('Yes')
# now check if frequency of all characters # can become same
Driver program
if name == "main": input = 'xxxyyzzt' allSame(input) ```
输出:
```py No
```
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