如何从Python中的字符串中提取数字？

Question

如何从字符串中提取数字以便能够操纵它？该号码可以是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

鉴于你的字符串的结构，当您使用str.split字符串分成三个字符串列表，你应该只取三个要素之一：

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

然后可以使用float该字符串转换成一个数字：

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

如果你不能像琴弦的结构，一定的，你可以使用re（正则表达式）中提取号码和map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

你试过的尝试，除了在你的类型强制转换模块将扔掉串粉，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

收件您类似下面它试图将其参数的第一转换为int，然后进入一个float，然后进入一个complex（只是延伸的例子）的一个有点转换功能。如果您希望获得/保留最适合输入的类型，那么尝试转换的顺序很重要，因为int将成功转换为float，但反之亦然，因此您需要尝试将输入转换为首先是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'>) 

你可以使用正则表达式剜出从输入的每一行的数字字段按jonrsharpe的答案。

Answer 4

从字符串中提取数字有一个非常简单和最好的方法。您可以使用以下代码从字符串中提取N位数字。

- 获得整数 -

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

-GET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))