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I am a newbie using Python and Numpy. I thought this would be simple and probably is. I have an array of times.
For example:

times = (0.5,  0.75,  1.5)

This array will vary in size depending on the files loaded.

I simply want to find the difference in time between each subsequent element. 

0.75 - 0.5
then
1.5 - 0.75

and so for the number of elements in the array. Then I put each result into one column.

I have tried various for loops but unable to do it. There must be a simple way?

Thanks, Scott

Levon
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Scott
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  • Well, what *have* you tried? Incidentally, this is a common problem. Search on here or on google. – Marcin May 29 '12 at 18:52
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    possible duplicate of [Python - Differences between elements of a list](http://stackoverflow.com/questions/2400840/python-differences-between-elements-of-a-list) – Marcin May 29 '12 at 18:54
  • Also http://stackoverflow.com/questions/5314241/difference-between-consecutive-elements-in-list – Marcin May 29 '12 at 18:54
  • Thanks Marcin, everything I looked at online or tried didn't answer this case (bad searching I guess). The trouble I had was getting indiv. elements to do calculations on each other over an array. Everything was about one element, not two. Your links are very helpful as are the other suggestions here. Thanks! – Scott May 29 '12 at 19:55

6 Answers6

7

How about this?

>>> import numpy as np
>>> a = np.array([0.5,  0.75,  1.5])
>>> np.diff(a)
array([ 0.25,  0.75])
>>>
Nick Craig-Wood
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4

First, note that for most everyday uses, you probably won't require an array (which is a special datatype in Numpy). Python's workhorse means of data storage is the list, and is definitely worth reading up on if you're coming from a more rigid programming language. Lists are actually defined using square brackets:

times = [ 0.5,  0.75,  1.5 ]

Then, with special syntax called a list comprehension, we can create a new list that has (length-1) elements. This expression automatically figures out the size of the list needed.

diffs = [ times[i] - times[i-1] for i in range(1, len(times)) ]

And for the sample data provided, returns:

[0.25, 0.75]
abought
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1

Or how about these (all variants on the same theme):

import numpy as np
a = np.array([0.5,  0.75,  1.5])
b = a[1:]-a[:-1]
print (b)

or without numpy:

a=[0.5,  0.75,  1.5]
a1=a[1:]
a2=a[:-1]
b=[ aa-a2[i] for i,aa in enumerate(a1) ]

or

a=[0.5,  0.75,  1.5]
c=[x-y for x,y in zip(a[1:],a[:-1])]
mgilson
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1
diffs = [b-a for a, b in zip(times[:-1], times[1:])]
[0.25, 0.75]

This approach requires no numpy, simple Python. Subtracting 

times[1:]
(0.75, 1.5)

times[:-1]
(0.5, 0.75)

from each other

Levon
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0

this should do it.

import numpy as np


times = np.array([0.5,0.75,1.5,2.0])
diff_times = np.zeros(len(times)-1,dtype =float)

for i in range(1,len(times)):
    diff_times[i-1] = times[i] - times[i-1]

print diff_times
fhtuft
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0

very basic:

liste = [1,3,8]
difference = []
for i in range(len(liste)-1):
    diffs = abs(liste[i] - liste[i+1])
    difference.append(diffs)

print difference
Gregor
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