如何从 Python 中的字符串中仅提取 integer 部分？

Question

我想只提取字符串中包含的数字。 可以为此目的组合isdigit()和split()还是有更简单/更快的方法？

例子：

m = ['How to extract only number 122', 'The number 35 must be extracted', '1052 must be extracted']

Output：

numbers = [122, 35, 1052]
text = ['How to extract only number', 'The number must be extracted', 'must be extracted']

我的代码：

text = []
numbers = []
temp_numbers = []
for i in range(len(m)):
    text.append([word for word in m[i].split() if not word.isdigit()])
    temp_numbers.append([int(word) for word in m[i].split() if word.isdigit()])
for i in range(len(m)):
    text[i] = ' '.join(text[i])
for elem in temp_numbers:
    numbers.extend(elem)

print(text)
print(numbers)

Answer 1

导入正则表达式库：

import re

如果要提取所有数字：

numbers = []
texts = []
for string in m:
    numbers.append(re.findall("\d+", string))
    texts.append(re.sub("\d+", "", string).strip())

如果您只想提取第一个数字：

numbers = []
texts = []
for string in m:
    numbers.append(re.findall("\d+", string)[0])
    texts.append(re.sub("\d+", "", string).strip())

Answer 2

因此，如果我们将 m 作为列表，您可以遍历它并检查当前字符是否为数字，如果是，则为 append。

For循环解决方案：

m = ['How to extract only number 122', 'The number 35 must be extracted', '1052 must be extracted']

numbers = []
temp_num = ""

for string in m:
    # Presuming m only contains strings

    for char in string:
        if char.isdigit():
            temp_num += char
    
    numbers.append(int(temp_num))
    temp_num = ""

列表理解解决方案 - 在不同的索引处附加每个数字：

m = ['How to extract only number 122', 'The number 35 must be extracted', '1052 must be extracted']

numbers = [int(char) for string in m for char in string if char.isdigit()]

希望这会有所帮助，如果您只想获取可迭代的值（例如列表），只需for varname in iterable它会更快、更清晰。

如果您需要索引和值，请使用for index, varname in enumerate(iterable) 。

Answer 3

nums_list = []
m = ["How to extract only number 122", "The number 35 must be extracted", "1052 must be extracted"]
for i in m:
    new_l = i.split(" ")
    for j in new_l:
        if j.isdigit():
            nums_list.append(int(j))
print nums_list

操作：

[122, 35, 1052]