Python从字符串中打印前N个整数

Question

Is it possible without regex in python to print the first n integers from a string containing both integers and characters?

For instance:

string1 = 'test120202test34234e23424'
string2 = 'ex120202test34234e23424'

foo(string1,6)  => 120202
foo(string2,6) => 120202

Answer 1

Anything's possible without a regex. Most things are preferable without a regex.

On easy way is.

>>> str = 'test120202test34234e23424'
>>> str2 = 'ex120202test34234e23424'
>>> ''.join(c for c in str if c.isdigit())[:6]
'120202'
>>> ''.join(c for c in str2 if c.isdigit())[:6]
'120202'

You might want to handle your corner cases some specific way -- it all depends on what you know your code should do.

>>> str3 = "hello 4 world"
>>> ''.join(c for c in str3 if c.isdigit())[:6]
'4'

And don't name your strings str !

Answer 2

You can remove all the alphabets from you string with str.translate and the slice till the number of digits you want, like this

import string

def foo(input_string, num):
    return input_string.translate(None, string.letters)[:num]

print foo('test120202test34234e23424', 6)   # 120202
print foo('ex120202test34234e23424',   6)   # 120202

Note: This simple technique works only in Python 2.x

But the most efficient way is to go with the itertools.islice

from itertools import islice

def foo(input_string, num):
    return "".join(islice((char for char in input_string if char.isdigit()),num))

This is is the most efficient way because, it doesn't have to process the entire string before returning the result.

Answer 3

If you didn't want to process the whole string - not a problem with the length of strings you give as an example - you could try:

import itertools
"".join(itertools.islice((c for c in str2 if c.isdigit()),0,5))