Question

如何计算文本中的空格数或新行字符数，使连续空格仅计为一个？例如，这非常接近我想要的：

string = "This is an  example text.\n   But would be good if it worked."
counter = 0
for i in string:
    if i == ' ' or i == '\n':
        counter += 1
print(counter)

但是，结果应该仅为15，而不是返回11。

Answer 1

假设您被允许使用Python正则表达式;

import re
print len(re.findall(ur"[ \n]+", string))

快捷方便！

更新：此外，使用[\s]代替[ \n]来匹配任何空格字符。

Answer 2

你可以这样做：

string = "This is an  example text.\n   But would be good if it worked."
counter = 0
# A boolean flag indicating whether the previous character was a space
previous = False 
for i in string:
    if i == ' ' or i == '\n': 
        # The current character is a space
        previous = True # Setup for the next iteration
    else:
        # The current character is not a space, check if the previous one was
        if previous:
            counter += 1

        previous = False
print(counter)

Answer 3

您可以遍历数字以将它们用作索引。

for i in range(1, len(string)):
    if string[i] in ' \n' and string[i-1] not in ' \n':
        counter += 1
if string[0] in ' \n':
    counter += 1
print(counter)

注意第一个符号，因为此构造从第二个符号开始，以防止IndexError。

Answer 4

默认的str.split()函数会将连续的空格运行视为一个空格。因此，只需拆分字符串，获取结果列表的大小，然后减去一个。

len(string.split())-1

Answer 5

你可以使用枚举，检查下一个char也不是空格，所以连续的空格只会算作1：

string = "This is an  example text.\n   But would be good if it worked."

print(sum(ch.isspace() and not string[i:i+1].isspace() for i, ch in enumerate(string, 1)))

您还可以将iter与生成器函数一起使用，跟踪最后一个字符并进行比较：

def con(s):
    it = iter(s)
    prev = next(it)
    for ele in it:
        yield prev.isspace() and not ele.isspace()
        prev = ele
    yield ele.isspace()

print(sum(con(string)))

itertools版本：

string = "This is an  example text.\n     But would be good if it worked.  "

from itertools import tee, izip_longest

a, b = tee(string)
next(b)
print(sum(a.isspace() and not b.isspace() for a,b in izip_longest(a,b, fillvalue="") ))

Answer 6

尝试：

def word_count(my_string):     
    word_count = 1
    for i in range(1, len(my_string)):
        if my_string[i] == " ":

            if not my_string[i - 1] == " ":    
                word_count += 1

         return word_count

Answer 7

您可以使用函数xmlns:app="http://schemas.android.com/apk/res-auto"查找连续的空格组：

groupby()

计算文本中的空格（将连续的空格视为一个）

7 个答案: