I did not try the second solution from Chris since it is too slow as Chris
> stated.
>
> Frank
>
>
> > Date: Thu, 2 Oct 2008 17:43:37 +0200
> > From: [EMAIL PROTECTED]
> > To: numpy-discussion@scipy.org
> > CC: [EMAIL PROTECTED]
> > Subject: Re: [Numpy-d
Oct 2008 17:43:37 +0200> From: [EMAIL PROTECTED]> To:
> numpy-discussion@scipy.org> CC: [EMAIL PROTECTED]> Subject: Re:
> [Numpy-discussion] Help to process a large data file> > Frank,> > I would
> imagine that you cannot get a much better performance in python
Frank,
I would imagine that you cannot get a much better performance in python
than this, which avoids string conversions:
c = []
count = 0
for line in open('foo'):
if line == '1 1\n':
c.append(count)
count = 0
else:
if '1' in line: count += 1
One could do some n
Frank,
How about that:
x = np.loadtxt('file')
z = x.sum(1) # Reduce data to an array of 0,1,2
rz = z[z>0] # Remove all 0s since you don't want to count those.
loc = np.where(rz==2)[0] # The location of the (1,1)s
count = np.diff(loc) - 1 # The spacing between those (1,1)s, ie, the numbe
Hi,
I have a large data file which contains 2 columns of data. The two columns only
have zero and one. Now I want to cound how many one in between if both columns
are one. For example, if my data is:
1 0
0 0
1 1
0 0
0 1x
0 1x
0 0
0 1x
1 1
0 0
0 1x
0 1x
1 1
Then my
