Reading fortran double precision format into python

Question

I am trying to read a Fortran double-precision number like 1.2345D+02 into python, but I got the following error:

>>> float('1.2345D+02')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for float(): 1.2345D+02

By following the advice on Python scientific notation using D instead of E, I tried numpy but I also get the same error:

import numpy
>>> numpy.float("1.2345D+02")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for float(): 1.2345D+02

Is there a solution in Python to read those double precision numbers without just changing the 'D' to 'E'?

EDIT: I replaced a bad syntax on the strings. But still I get errors.

Just a heads up that maybe the fortranformat Python library (on PyPI) can help - I should probably disclose that I'm the author. — Brendan
– Brendan, Commented Jun 29, 2011 at 18:32

orlp · Accepted Answer · 2011-05-13 08:42:22Z

5

What's wrong with float(str.replace("D", "E"))?

Please note, numpy DOES support fortran notation: numpy.float("1.2345D+02").

You seem to have some deeper purpose, perhaps shedding light on it would help.

answered May 13, 2011 at 8:42

orlp

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9 Comments

iluvatar Over a year ago

In general, nothing is wrong with replacing "D" with "E". I just want to know if there is a "native" solution. If numpy supports fortran notation, why is it failing as shown in the example? Yes, I need to parse some huge fortran-generated data files and to extract some particular data. Thanks!

orlp Over a year ago

Because numpy.float(1.2345D+02) would be parsed by the python interpreter as a call to numpy.float with 1.2345D+02. 1.2345D+02 is unknown to the python interpreter so it gives an error. "1.2345D+02" is a string, which is perfectly fine.

iluvatar Over a year ago

Thanks. Actually it was a typo on my example. I already corrected it but the error (another one) remains.

John La Rooy Over a year ago

Possibly you are using different versions of numpy

jfs Over a year ago

At least numpy 1.5.1 doesn't support numpy.float("1.2345D+02").

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Mike T · Accepted Answer · 2012-05-16 02:53:31Z

1

The substitution can be made a bit more careful by using a regular expression:

import re
re_dbl_fort = re.compile(r'(\d*\.\d+)[dD]([-+]?\d+)')

text = 'DEW=[0.242D+03 -4.320D-06]'
re_dbl_fort.sub(r'\1E\2', text)
# DEW=[0.242E+03 -4.320E-06]

Or if you have a list of strings (lines) read from a file using readlines():

lines = ['REPORT CARD\n', 'GRADE: D+ (1.3D+00/4.0D+00)\n']
for ln, line in enumerate(lines):
    res = re_dbl_fort.sub(r'\1E\2', line)
    if line != res:
        lines[ln] = res
# ['REPORT CARD\n', 'GRADE: D+ (1.3E+00/4.0E+00)\n']

answered May 16, 2012 at 2:53

Mike T

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