计算字符串中的字母频率 (Python)
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Counting Letter Frequency in a String (Python)
提问by Kelvin San
I am trying to count the occurrences of each letter of a word
我正在尝试计算单词的每个字母的出现次数
word = input("Enter a word")
Alphabet=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
for i in range(0,26):
print(word.count(Alphabet[i]))
This currently outputs the number of times each letter occurs including the ones that don't.
这当前输出每个字母出现的次数，包括没有出现的次数。
How do I list the letters vertically with the frequency alongside it e.g:
我如何垂直列出字母及其旁边的频率，例如：
word="Hello"
字=“你好”
H 1
1
E 1
1
L 2
2
O 1
1
回答by LMc
from collections import Counter
counts=Counter(word) # Counter({'l': 2, 'H': 1, 'e': 1, 'o': 1})
for i in word:
print(i,counts[i])
Try using Counter, which will create a dictionary that contains the frequencies of all items in a collection.
尝试使用Counter，这将创建一个字典，其中包含集合中所有项目的频率。
Otherwise, you could do a condition on your current code to printonly if word.count(Alphabet[i])is greater than 0, though that would slower.
否则，您可以对当前代码设置条件，print仅当word.count(Alphabet[i])大于 0 时，尽管这会变慢。
回答by Uzam Hashmi
def char_frequency(str1):
dict = {}
for n in str1:
keys = dict.keys()
if n in keys:
dict[n] += 1
else:
dict[n] = 1
return dict
print(char_frequency('google.com'))
回答by duhaime
As @Pythonista said, this is a job for collections.Counter:
正如@Pythonista 所说，这是一份工作collections.Counter：
from collections import Counter
print(Counter('cats on wheels'))
This prints:
这打印：
{'s': 2, ' ': 2, 'e': 2, 't': 1, 'n': 1, 'l': 1, 'a': 1, 'c': 1, 'w': 1, 'h': 1, 'o': 1}
回答by soheshdoshi
easy and simple solution without lib.
没有lib的简单简单的解决方案。
string=input()
f={}
for i in string:
f[i]=f.get(i,0)+1
print(f)
here is the link for get()https://docs.quantifiedcode.com/python-anti-patterns/correctness/not_using_get_to_return_a_default_value_from_a_dictionary.html
回答by Shahriar Rahman Zahin
If using libraries or built-in functions is to be avoided then the following code may help:
如果要避免使用库或内置函数，那么以下代码可能会有所帮助：
s = "aaabbc" # sample string
dict_counter = {} # empty dict for holding characters as keys and count as values
for char in s: # traversing the whole string character by character
if not dict_counter or char not in dict_counter.keys(): # Checking whether the dict is
# empty or contains the character
dict_counter.update({char: 1}) # if not then adding the character to dict with count = 1
elif char in dict_counter.keys(): # if the char is already in the dict then update count
dict_counter[char] += 1
for key, val in dict_counter.items(): # Looping over each key and value pair for printing
print(key, val)
Output:
a 3
b 2
c 1
输出：
a 3
b 2
c 1
回答by Swetank
s=input()
t=s.lower()
for i in range(len(s)):
b=t.count(t[i])
print("{} -- {}".format(s[i],b))
回答by Syed Rafay
Another way could be to remove repeated characters and iterate only on the unique characters (by using set()) and then counting the occurrence of each unique character (by using str.count())
另一种方法可能是删除重复字符并仅迭代唯一字符（使用set()），然后计算每个唯一字符的出现次数（使用str.count()）
def char_count(string):
freq = {}
for char in set(string):
freq[char] = string.count(char)
return freq
if __name__ == "__main__":
s = "HelloWorldHello"
print(char_count(s))
# Output: {'e': 2, 'o': 3, 'W': 1, 'r': 1, 'd': 1, 'l': 5, 'H': 2}
回答by Stanley
For future references: When you have a list with all the words you want, lets say
wordlistit's pretty simple
供以后参考：当你有一个包含你想要的所有单词的列表时，让我们说
wordlist这很简单
for numbers in range(len(wordlist)):
if wordlist[numbers][0] == 'a':
print(wordlist[numbers])
回答by evadeflow
Following up what LMc said, your code was already pretty close to functional, you just needed to post-process the result set to remove 'uninteresting' output. Here's one way to make your code work:
按照 LMc 所说，您的代码已经非常接近功能，您只需要对结果集进行后处理以删除“无趣”的输出。这是使您的代码工作的一种方法：
#!/usr/bin/env python
word = raw_input("Enter a word: ")
Alphabet = [
'a','b','c','d','e','f','g','h','i','j','k','l','m',
'n','o','p','q','r','s','t','u','v','w','x','y','z'
]
hits = [
(Alphabet[i], word.count(Alphabet[i]))
for i in range(len(Alphabet))
if word.count(Alphabet[i])
]
for letter, frequency in hits:
print letter.upper(), frequency
But the solution using collections.Counteris much more elegant/Pythonic.
但是使用的解决方案collections.Counter更加优雅/Pythonic。
