Integer pattern - Python regex -

I have seen many posts on this but I still can not work it, I do not know why.

I have some floating point and it has a relatively simple string with integer numbers such as: '2 1.000000000000000 1 1 0' . I want to extract only the integer from it, in this example only 2, 1, 1, 0 ( 1 is not that which is 0 S) .

I know that I will have to use the numbers that use lookbehind and lookahead to exclude Code>. I can successfully find numbers that occur before a coma, in that case 0 :

  import again fullCertain = recompackage ('- (? ? & Lt; = \.) \ D + (?!?) ') A =' 2 1.00000000000000000 1 1 0 'FullPortal Fondol (A)  

Back to ['000000000000000'] , as I want but when I try to find the numbers which no < / Em> before . s does not work:

  Re-import IntegerPattern = re Compile ('- (?? (? & Lt ;! \.) \ D + (?! \ ) 'A' = '2 1.000000000000000 1 1' 'Integerparent Tern. Fundol (A)  

Return [' 2 ',' 00000000000000 ',' 1 ',' 1 ' '0'] . Any ideas why? I am completely new to general expression and it just does not leave me. This should work but it is not.

  (\ S | ^) \ d + (\ s | $)  

The code can be

  & Gt; & Gt; & Gt; N = '2 1.000000000000000 1 1 0' & gt; & Gt; & Gt; ('? (?: \ S | ^) \ d + (?: \ S | $)', n) ['2', '1', '0']  

(\ s * | ^) matches a location or the beginning of the string

\ d + matches any number of digits

(\ s * | $) matches the end of the location or string

Note: \ b to < Code> \ d + / Code> as . also \ b

example

edit

why include Regex (? & Lt ;! \.) \ D + (?!!) work

The problem here is now that when the argument seems negative, then we match the match . and regex .

When you type (? regex is a condition where it can be successful

He is saying 1.000000 regex condition fixes second 0 , as the previous state is not . (which is zero) and the remaining 00000 is thus a winner. Therefore it matches this.

Check this link to get a clear view

As you 1.00000000000000000 match

1

A more correct regex will be

(? :(? & Lt; = ^) | (? & Lt; = \ s)) \ d + (? = \ S | $)

  & Gt; & Gt; N '1 2 3 4.5' & gt; & Gt; & Gt; Re.findall (r '(?: (? & Lt; = ^) | (? & Lt; = \ s)) \ d + (? = \ S | $)', n) ['1', '2 ',' 3 ']> gt; & Gt; & Gt; N = '1 2 3 4' & gt; & Gt; & Gt; Re.findall (r '(?: (? & Lt; = ^) | (? & Lt; = \ s)) \ d + (? = \ S | $)', n) ['1', '2 Thank you for telling you