str information

[nicstel]

How to print only the number in:

x = 1p
if I print x = 1p
but in want only = 1

Thank You

[boxfish]

I'm guessing you mean x = "1p"?
You're trying to take an integer value off the front of a string, right?
So if x = "123abc" you want 123, and if x = "10307a10c5 2" you want 10307, right? I don't know of a function that does this for you in Python, but I can think of some code that would, although it probably isn't very efficient code. You could make a loop that takes larger portions of the string each iteration, tries to convert them to integers with the int() function, and ends when it catches a ValueError. Here it is, for what it's worth:

x = "123abc"
x_number = 0
for i in xrange(1, len(x)):
    try:
        x_number = int(x[:i])
    except ValueError:
        break
print x_number

I also googled for "Python, atoi", and found this archived thread, in which reply #3 has some code that does this too.

Hope this helps.

[bvdet]

Below are three ways of getting the numbers from a string. The first one gets all the numbers, and the other two get only the leading numbers.[code=Python]x = "123abc456"
print ''.join([c for c in x if c.isdigit()])

import re
patt = re.compile(r'^([0-9]+)')
print patt.search(x). group(1)

x_num = ''
for c in x:
if c.isdigit():
x_num += c
else:
break
print x_num[/code]Output:
>>> 123456
123
123

[nicstel]

It was exactly what I wanted

Thank you at both of you

[ghostdog74]

KISS

>>> import string
>>> x="abc123"
>>> x.strip(string.letters)
'123'
>>> x="123abc"
>>> x.strip(string.letters)
'123'
>>>

[nicstel]

Thank ghostdog74,

It's more simple. Have a nice day.