我正在尝试计算单词中每个字母出现的次数
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]))
当前代码输出每个字母出现的次数，包括那些没有出现过的。
如何垂直列出每个字母及其频率，例如下面这样？
word="Hello"
H 1
E 1
L 2
O 1
from collections import Counter
counts=Counter(word) # Counter({'l': 2, 'H': 1, 'e': 1, 'o': 1})
for i in word:
print(i,counts[i])
尝试使用Counter，它将创建一个字典，其中包含集合中所有项目的频率。
否则，您可以对当前代码进行条件判断，仅在word.count(Alphabet[i])大于0时print，但这会更慢。
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'))
正如Pythonista所说的那样，这是collections.Counter的工作：
from collections import Counter
print(Counter('cats on wheels'))
这将打印：
{'s': 2, ' ': 2, 'e': 2, 't': 1, 'n': 1, 'l': 1, 'a': 1, 'c': 1, 'w': 1, 'h': 1, 'o': 1}
一种不需要使用外部库的简单易行的解决方案：
string = input()
f = {}
for i in string:
f[i] = f.get(i,0) + 1
print(f)
get()的链接: https://docs.quantifiedcode.com/python-anti-patterns/correctness/not_using_get_to_return_a_default_value_from_a_dictionary.html，该链接解释了为什么应该使用get()来从字典中返回默认值。s = input()
t = s.lower()
for i in range(len(s)):
b = t.count(t[i])
print("{} -- {}".format(s[i], b))
针对LMc所说的内容, 您的代码已经非常接近于功能实现。您只需要对结果集进行后处理，以去除“不相关”的输出。下面是让您的代码正常工作的一种方法：
#!/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
然而使用 collections.Counter 的解决方案更加优雅/Pythonic。
在编程中，包含所有字母可能是有意义的。例如，如果您想计算单词分布之间的余弦差异，通常需要所有字母。
您可以使用以下方法：
from collections import Counter
def character_distribution_of_string(pass_string):
letters = ["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"]
chars_in_string = Counter(pass_string)
res = {}
for letter in letters:
if(letter in chars_in_string):
res[letter] = chars_in_string[letter]
else:
res[letter] = 0
return(res)
使用方法：
character_distribution_of_string("This is a string that I want to know about")
完整字符分布
{'a': 4,
'b': 1,
'c': 0,
'd': 0,
'e': 0,
'f': 0,
'g': 1,
'h': 2,
'i': 3,
'j': 0,
'k': 1,
'l': 0,
'm': 0,
'n': 3,
'o': 3,
'p': 0,
'q': 0,
'r': 1,
's': 3,
't': 6,
'u': 1,
'v': 0,
'w': 2,
'x': 0,
'y': 0,
'z': 0}
list(character_distribution_of_string("This is a string that I want to know about").values())
giving...
[4, 1, 0, 0, 0, 0, 1, 2, 3, 0, 1, 0, 0, 3, 3, 0, 0, 1, 3, 6, 1, 0, 2, 0, 0, 0]
wordlist，它就非常简单了。for numbers in range(len(wordlist)):
if wordlist[numbers][0] == 'a':
print(wordlist[numbers])
import string
word = input("Enter a word: ")
word = word.lower()
Alphabet=list(string.ascii_lowercase)
res = []
for i in range(0,26):
res.append(word.count(Alphabet[i]))
for i in range (0,26):
if res[i] != 0:
print(str(Alphabet[i].upper()) + " " + str(res[i]))
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 character 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)
输出：
a 3
b 2
c 1
collections.Counter。 - Pythonista