1

我需要从字符串中分离出所有整数，列出它们，并打印它们的总和和整数的#。我的问题是我当前的代码会将例如 456 拆分为 4、5 和 6，并将它们视为单独的整数。不幸的是，正则表达式不是一种选择。

到目前为止我所拥有的：

def tally(text):
s = ','.join(x for x in text if x.isdigit())
numbers = [int(x) for x in s.split(",")]
num=len(numbers)
t=sum(numbers)
print ('There are', num, 'integers in the input summing up to', t)

. 

What i need: input:'34 ch33se 34e8 3.4'
output: [34 33 34 8 3 4 ]
im getting now is [3 4 3 3 8 3 4]
4

4 回答 4

1

这是一个单行：

>>> s = '34 ch33se 34e8 3.4'
>>> map(int, filter(None, ''.join(map(lambda c: (c.isdigit() and c or ' '), s)).split(' ')))
[34, 33, 34, 8, 3, 4]
于 2013-09-15T23:38:48.320 回答
0

使用的选项itertools.groupby：

>>> from itertools import groupby
>>> s = '34 ch33se 34e8 3.4'
>>>
>>> filter(str.isdigit, (''.join(g) for k,g in groupby(s, key=str.isdigit)))
<filter object at 0x1006ebc50>
>>> list(_)
['34', '33', '34', '8', '3', '4']

这比公认的答案要快得多：

>>> from timeit import timeit
>>> 
>>> setup = """
... from itertools import groupby
... s = '34 ch33se 34e8 3.4'
... """
>>> 
>>> timeit("map(int, filter(None, ''.join(map(lambda c: (c.isdigit() and c or ' '), s)).split(' ')))", setup)
6.1007173559919465
>>> timeit("filter(str.isdigit, (''.join(g) for k,g in groupby(s, key=str.isdigit)))", setup)
1.5044781469914597
于 2013-09-15T23:51:58.503 回答
0 
def get_integers(string):
    current_integer = ""
    for i in string:
        if i.isdigit():
            current_integer += str(i)
        else:
            if current_integer:
                yield int(current_integer)
                current_integer = ""
    if current_integer:
        yield int(current_integer)

# Prints [34, 33, 34, 8, 3, 4]
print([i for i in get_integers('34 ch33se 34e8 3.4')])

你应该能够sum或任何结果。

于 2013-09-15T23:36:50.867 回答
0

为什么不创建一个逐字符搜索字符串并获取数字的函数？像这样的东西：

def find_numbers(string):
    numbers = []
    usable = True #mark unusable if its inside a word
    latest_number = ''
    for char in string:
        if char in '1234567890' and usable:
            latest_number += char
            continue
        elif latest_number:
            numbers.append(int(latest_number))
            latest_number = ''
        if char in ' ,':
            usable = True
        else:
            usable = False
    if latest_number: numbers.append(int(latest_number))
    return numbers

使用此功能，您可以轻松获取数字：

>>> a = 'hello 123 wh3at? 2'
>>> find_numbers(a)
[123, 2]

编辑：

如果要包含单词内的整数：

def find_numbers(string):
    numbers = []
    latest_number = ''
    for char in string:
        if char in '1234567890':
            latest_number += char
        elif latest_number:
            numbers.append(int(latest_number))
            latest_number = ''
    if latest_number: numbers.append(int(latest_number))
    return numbers
于 2013-09-15T23:45:49.160 回答