如何计算字符串中字符出现的次数?
如。“a”在“Mary had a little lamb”中出现了4次。
如何计算字符串中字符出现的次数?
如。“a”在“Mary had a little lamb”中出现了4次。
当前回答
你可以使用.count():
>>> 'Mary had a little lamb'.count('a')
4
其他回答
a = 'have a nice day'
symbol = 'abcdefghijklmnopqrstuvwxyz'
for key in symbol:
print(key, a.count(key))
这是公认答案的延伸,你应该在文本中寻找所有字符的计数。
# Objective: we will only count for non-empty characters
text = "count a character occurrence"
unique_letters = set(text)
result = dict((x, text.count(x)) for x in unique_letters if x.strip())
print(result)
# {'a': 3, 'c': 6, 'e': 3, 'u': 2, 'n': 2, 't': 2, 'r': 3, 'h': 1, 'o': 2}
我不知道“最简单的”,但简单的理解可以做到:
>>> my_string = "Mary had a little lamb"
>>> sum(char == 'a' for char in my_string)
4
利用内置的和,生成器理解和bool是整数的子类的事实:如何乘字符等于'a'。
要查找句子中字符的出现情况,您可以使用下面的代码
首先,我从句子中取出了唯一的字符,然后我计算了每个字符在句子中的出现次数,其中包括空格的出现次数。
ab = set("Mary had a little lamb")
test_str = "Mary had a little lamb"
for i in ab:
counter = test_str.count(i)
if i == ' ':
i = 'Space'
print(counter, i)
以上代码的输出如下所示。
1 : r ,
1 : h ,
1 : e ,
1 : M ,
4 : a ,
1 : b ,
1 : d ,
2 : t ,
3 : l ,
1 : i ,
4 : Space ,
1 : y ,
1 : m ,
不使用Counter(), count和regex获得所有字符计数的另一种方法
counts_dict = {}
for c in list(sentence):
if c not in counts_dict:
counts_dict[c] = 0
counts_dict[c] += 1
for key, value in counts_dict.items():
print(key, value)