match - Use python re.findall to split the line

Question

I am trying to use re.findall to split one string:

string = '1.1 2 -4259.8774  0.000000  0.707664  0.002210 -0.004314-0.004912-0.000823'

I tried with:

match = re.findall(r'-?d+.?d+m?', string)

but I got:

['1.1', '-4259.8774', '0.000000', '0.707664', '0.002210', '-0.004314', '-0.004912',
 '-0.000823']

The second string '2' is missing. What I want is:

['1.1', '2',  '-4259.8774', '0.000000', '0.707664', '0.002210', '-0.004314', '-0.004912',
 '-0.000823']

Answer 1

I would use re.findall here:

string = '1.1 2 -4259.8774  0.000000  0.707664  0.002210 -0.004314-0.004912-0.000823'
nums = re.findall(r'(?:|-)d+(?:.d+)?', string)
print(nums)

This prints:

['1.1', '2', '-4259.8774', '0.000000', '0.707664', '0.002210', '-0.004314', '-0.004912',
 '-0.000823']

Here is an explanation of the regex pattern:

(?:|-)       match either a word boundary OR a minus sign, which is followed by
d+(?:.d+)?  a whole number with optional decimal component

The idea here is that the left boundary of each number is either a word boundary, or the number starts with a minus sign.