How can I count the number of characters at the start/end of a string in Python?
For example, if the string is
'ffffhuffh'
How would I count the number of f
s at the start of the string? The above string with a f
should output 4.
str.count
is not useful to me as a character could be in the middle of the string.
A short and simple way will be to use the str.lstrip
method, and count the difference of length.
s = 'ffffhuffh'
print(len(s)-len(s.lstrip('f')))
# output: 4
Return a copy of the string with leading characters removed. The chars argument is a string specifying the set of characters to be removed.
Try this, using itertools.takewhile() :
import itertools as it
s = 'ffffhuffh'
sum(1 for _ in it.takewhile(lambda c: c == 'f', s))
=> 4
Similarly, for counting the characters at the end:
s = 'huffhffff'
sum(1 for _ in it.takewhile(lambda c: c == 'f', reversed(s)))
=> 4
You may use regular expression with re.match
to find the occurrence of any character at the start of the string as:
>>> import re
>>> my_str = 'ffffhuffh'
>>> my_char = 'f'
>>> len(re.match('{}*'.format(my_char), my_str).group())
4
Building on Oscar Lopez's answer, I want to handle the case you mention of the end of the string: use reversed()
import itertools as it
my_string = 'ffffhuffh'
len(list(it.takewhile(lambda c: c == my_string[-1], reversed(my_string))))
=> 1
You can create a function and iterate through your string and return the count of the desired char in the input string's beginning or end like this example:
# start = True: Count the chars in the beginning of the string
# start = False: Count the chars in the end of the string
def count_char(string= '', char='', start=True):
count = 0
if not start:
string = string[::-1]
for k in string:
if k is char:
count += 1
else:
break
return count
a = 'ffffhuffh'
print(count_char(a, 'f'))
b = a[::-1]
print(count_char(b, 'f', start=False))
Output:
4
4
You may also use itertools.groupby
to find the count of the occurrence of the first element at the start of the string as:
from itertools import groupby
def get_first_char_count(my_str):
return len([list(j) for _, j in groupby(my_str)][0])
Sample run:
>>> get_first_char_count('ffffhuffh')
4
>>> get_first_char_count('aywsnsb')
1
re.sub选择带重复的第一个字母((^(\\ w)\\ 2 *)),len计数频率。
len(re.sub(r'((^\w)\2*).*',r'\1',my_string))
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