如何在Python中从字符串中提取数字？

Question

如何从字符串中提取数字以进行操作？ 该数字可以是int或float 。 例如，如果字符串是"flour, 100, grams"或"flour, 100.5, grams"则提取数字100或100.5 。

代码 ：

string  = "flour, 100, grams"
numbers = [int(x) for x in string.split(",")]
print(numbers)

输出 ：

Traceback (most recent call last):
  File "/Users/lewis/Documents/extracting numbers.py", line 2, in <module>
    numbers = [int(x) for x in string.split(",")]
 File "/Users/lewis/Documents/extracting numbers.py", line 2, in <listcomp>
   numbers = [int(x) for x in string.split(",")]
ValueError: invalid literal for int() with base 10: 'flour'

Answer 1

给定字符串的结构，当您使用str.split将字符串拆分为三个字符串的列表时，您应该只采用以下三个元素之一：

>>> s = "flour, 100, grams"
>>> s.split(",")
['flour', ' 100', ' grams']
>>> s.split(",")[1] # index the middle element (Python is zero-based)
' 100'

然后，您可以使用float将字符串转换为数字：

>>> float(s.split(",")[1])
100.0

如果不确定字符串的结构，可以使用re （正则表达式）提取数字并map以将它们全部转换：

>>> import re
>>> map(float, re.findall(r"""\d+ # one or more digits
                              (?: # followed by...
                                  \. # a decimal point 
                                  \d+ # and another set of one or more digits
                              )? # zero or one times""",
                          "Numbers like 1.1, 2, 34 and 15.16.",
                          re.VERBOSE))
[1.1, 2.0, 34.0, 15.16]

Answer 2

您是否尝试过除铸型周围的块以外的其他块，这将扔掉细粉，但保持100

string = 'flour, 100, grams'
numbers = []

    for i in string.split(','):
    try:
        print int(i)
        numbers.append(i)
    except: pass

Answer 3

给自己写一个转换函数，就像下面的转换函数一样，它首先尝试将其参数转换为int ，然后转换为float ，然后转换为complex （只是扩展示例）。 如果您希望获取/保留最适合输入的类型，则尝试转换的顺序很重要，因为int将成功转换为float ，反之则不然，因此您需要尝试将输入转换为float首先是int 。

def convert_to_number(n):
    candidate_types = (int, float, complex)
    for t in candidate_types:
        try:
            return t(str(n))
        except ValueError:
#            pass
            print "{!r} is not {}".format(n, t)    # comment out if not debugging
    else:
        raise ValueError('{!r} can not be converted to any of: {}'.format(n, candidate_types))

>>> s = "flour, 100, grams"
>>> n = convert_to_number(s.split(',')[1])
>>> type(n)
<type 'int'>
>>> n
100

>>> s = "flour, 100.123, grams"
>>> n = convert_to_number(s.split(',')[1])
' 100.123' is not <type 'int'>
>>> type(n)
<type 'float'>
>>> n
100.123

>>> n = convert_to_number('100+20j')
'100+20j' is not <type 'int'>
'100+20j' is not <type 'float'>
>>> type(n)
<type 'complex'>
>>> n
(100+20j)

>>> n = convert_to_number('one')
'one' is not <type 'int'>
'one' is not <type 'float'>
'one' is not <type 'complex'>
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/tmp/ctn.py", line 10, in convert_to_number
    raise ValueError('{!r} can not be converted to any of: {}'.format(n, candidate_types))
ValueError: 'one' can not be converted to any of: (<type 'int'>, <type 'float'>, <type 'complex'>)

您可以使用正则表达式根据jonrsharpe的答案从输入的每一行中提取数字字段。

Answer 4

有一个非常简单和最佳的方法来从字符串中提取数字。 您可以使用以下代码从字符串中提取N个数字。

-获取整数-

import re
s = 'flour, 100, grams, 200HC'
print(re.findall('\d+', s))

-获取浮点数-

import re
map(float, re.findall(r"""\d+ # one or more digits
                          (?: # followed by...
                              \. # a decimal point 
                              \d+ # and another set of one or more digits
                          )? # zero or one times""",
                      "Numbers like 1.1, 2, 34 and 15.16.",
                      re.VERBOSE))