br
is the name of a list of strings that goes like this:
['14 0.000000 -- (long term 0.000000)\n',
'19 0.000000 -- (long term 0.000000)\n',
'22 0.000000 -- (long term 0.000000)\n',
...
I am interested in the first two columns, which I would like to convert to a numpy array. So far, I've come up with the following solution:
x = N.array ([0., 0.])
for i in br:
x = N.vstack ( (x, N.array (map (float, i.split ()[:2]))) )
This results into having a 2-D array:
array([[ 0., 0.],
[ 14., 0.],
开发者_运维知识库 [ 19., 0.],
[ 22., 0.],
...
However, since br
is rather big (~10^5 entries), this procedure takes some time.
I was wondering, is there a way to accomplish the same result, but in less time?
This is dramatically faster for me:
import numpy as N
br = ['14 0.000000 -- (long term 0.000000)\n']*50000
aa = N.zeros((len(br), 2))
for i,line in enumerate(br):
al, strs = aa[i], line.split(None, 2)[:2]
al[0], al[1] = float(strs[0]), float(strs[1])
Changes:
- Preallocate the numpy array (this is big). You already know you want a 2-dimensional array with particular dimensions.
- Only split() for the first 2 columns, since you don't want the rest.
- Don't use map(): it's slower than list comprehensions. I didn't even use list comprehensions, since you know you only have 2 columns.
- Assign directly into the preallocated array instead of generating new temp arrays as you iterate.
You can try to preprocess (with awk for exemple) the list of strings if they come from a file, and use numpy.fromtxt. If you can't do anything about the way you get this list, you have several possibilities:
- give up. You will run this function once a day. You don't care about speed, and your actual solution is good enough
- write an IO plugin with cython. You have a big potential gain because you will be able to do all the loops in c, and affects directly the values in a big (10^5, 2) numpy ndarray
- try another language to fix your problem. If using languages such as c or haskell, you may use ctypes to call the functions compiled in a dll from python
edit
maybe this approach is slightly faster:
def conv(mysrt):
return map(float, mystr.split()[:2])
br_float = map(conv, br)
x = N.array(br_float)
Changing
map (float, i.split()[:2])
to
map (float, i.split(' ',2)[:2])
might result in a slight speedup. Since you only care about first two space-separated items in each line there is no need to split the entire line. The 2
in i.split(' ',2)
tells split
to just make a maximum of 2 splits. For example,
In [11]: x='14 0.000000 -- (long term 0.000000)\n'
In [12]: x.split()
Out[12]: ['14', '0.000000', '--', '(long', 'term', '0.000000)']
In [13]: x.split(' ',2)
Out[13]: ['14', '0.000000', '-- (long term 0.000000)\n']
精彩评论