limit how much data is read with numpy.genfromtxt for matplotlib

Question

I am creating a graph in python using a text file for the source data and matplotlib to plot the graph. The simple logic below works well.

But is there a way to get

Answer 1

No idea about numpy, but one possible solution would be to use the stringio class.

That allows you to just load the data you actually need into a string with normal file IO (there's also a byte version), create a file-like object from the string and pass that to numpy.

Answer 2

numpy.genfromtxt accepts iterators as well as files. That means it will accept the output of itertools.islice. Here, test.txt is a five-line file:

>>> import itertools, numpy
>>> with open('test.txt') as t_in:
...     numpy.genfromtxt(itertools.islice(t_in, 3))
... 
array([[  1.,   2.,   3.,   4.,   5.],
       [  6.,   7.,   8.,   9.,  10.],
       [ 11.,  12.,  13.,  14.,  15.]])

One might think this would be slower than letting numpy handle the file IO, but a quick test suggests otherwise. genfromtxt provides a skip_footer keyword argument that you can use if you know how long the file is...

>>> numpy.genfromtxt('test.txt', skip_footer=2)
array([[  1.,   2.,   3.,   4.,   5.],
       [  6.,   7.,   8.,   9.,  10.],
       [ 11.,  12.,  13.,  14.,  15.]])

...but a few informal tests on a 1000-line file suggest that using islice is faster even if you skip only a few lines:

>>> def get(nlines, islice=itertools.islice):
...     with open('test.txt') as t_in:
...         numpy.genfromtxt(islice(t_in, nlines))
...         
>>> %timeit get(3)
1000 loops, best of 3: 338 us per loop
>>> %timeit numpy.genfromtxt('test.txt', skip_footer=997)
100 loops, best of 3: 4.92 ms per loop
>>> %timeit get(300)
100 loops, best of 3: 5.04 ms per loop
>>> %timeit numpy.genfromtxt('test.txt', skip_footer=700)
100 loops, best of 3: 8.48 ms per loop
>>> %timeit get(999)
100 loops, best of 3: 16.2 ms per loop
>>> %timeit numpy.genfromtxt('test.txt', skip_footer=1)
100 loops, best of 3: 16.7 ms per loop