Given a list of numbers, how does one find differences between every (i)-th elements and its (i+1)-th?

Is it better to use a lambda expression or maybe a list comprehension?

For example:

Given a list t=[1,3,6,...], the goal is to find a list v=[2,3,...] because 3-1=2, 6-3=3, etc.

>>> t
[1, 3, 6]
>>> [j-i for i, j in zip(t[:-1], t[1:])]  # or use itertools.izip in py2k
[2, 3]

The other answers are correct but if you’re doing numerical work, you might want to consider numpy. Using numpy, the answer is:

v = numpy.diff(t)

If you don’t want to use numpy nor zip, you can use the following solution:

>>> t = [1, 3, 6]
>>> v = [t[i+1]-t[i] for i in range(len(t)-1)]
>>> v
[2, 3]

You can use itertools.tee and zip to efficiently build the result:

from itertools import tee
# python2 only:
#from itertools import izip as zip

def differences(seq):
    iterable, copied = tee(seq)
    next(copied)
    for x, y in zip(iterable, copied):
        yield y - x

Or using itertools.islice instead:

from itertools import islice

def differences(seq):
    nexts = islice(seq, 1, None)
    for x, y in zip(seq, nexts):
        yield y - x

You can also avoid using the itertools module:

def differences(seq):
    iterable = iter(seq)
    prev = next(iterable)
    for element in iterable:
        yield element - prev
        prev = element

All these solution work in constant space if you don’t need to store all the results and support infinite iterables.

Here are some micro-benchmarks of the solutions:

In [12]: L = range(10**6)

In [13]: from collections import deque
In [15]: %timeit deque(differences_tee(L), maxlen=0)
10 loops, best of 3: 122 ms per loop

In [16]: %timeit deque(differences_islice(L), maxlen=0)
10 loops, best of 3: 127 ms per loop

In [17]: %timeit deque(differences_no_it(L), maxlen=0)
10 loops, best of 3: 89.9 ms per loop

And the other proposed solutions:

In [18]: %timeit [x[1] - x[0] for x in zip(L[1:], L)]
10 loops, best of 3: 163 ms per loop

In [19]: %timeit [L[i+1]-L[i] for i in range(len(L)-1)]
1 loops, best of 3: 395 ms per loop

In [20]: import numpy as np

In [21]: %timeit np.diff(L)
1 loops, best of 3: 479 ms per loop

In [35]: %%timeit
    ...: res = []
    ...: for i in range(len(L) - 1):
    ...:     res.append(L[i+1] - L[i])
    ...: 
1 loops, best of 3: 234 ms per loop

Note that:

• zip(L[1:], L) is equivalent to zip(L[1:], L[:-1]) since zip already terminates on the shortest input, however it avoids a whole copy of L.
• Accessing the single elements by index is very slow because every index access is a method call in python

• numpy.diff is slow because it has to first convert the list to a ndarray. Obviously if you start with an ndarray it will be much faster:

  In [22]: arr = np.array(L)
  
  In [23]: %timeit np.diff(arr)
  100 loops, best of 3: 3.02 ms per loop
  

Using the := walrus operator available in Python 3.8+:

>>> t = [1, 3, 6]
>>> prev = t[0]; [-prev + (prev := x) for x in t[1:]]
[2, 3]

I would suggest using

v = np.diff(t)

this is simple and easy to read.

But if you want v to have the same length as t then

v = np.diff([t[0]] + t) # for python 3.x

or

v = np.diff(t + [t[-1]])

FYI: this will only work for lists.

for numpy arrays

v = np.diff(np.append(t[0], t))

A functional approach:

>>> import operator
>>> a = [1,3,5,7,11,13,17,21]
>>> map(operator.sub, a[1:], a[:-1])
[2, 2, 2, 4, 2, 4, 4]

Using generator:

>>> import operator, itertools
>>> g1,g2 = itertools.tee((x*x for x in xrange(5)),2)
>>> list(itertools.imap(operator.sub, itertools.islice(g1,1,None), g2))
[1, 3, 5, 7]

Using indices:

>>> [a[i+1]-a[i] for i in xrange(len(a)-1)]
[2, 2, 2, 4, 2, 4, 4]

Ok. I think I found the proper solution:

v = [x[1]-x[0] for x in zip(t[1:],t[:-1])]

Solution with periodic boundaries

Sometimes with numerical integration you will want to difference a list with periodic boundary conditions (so the first element calculates the difference to the last. In this case the numpy.roll function is helpful:

v-np.roll(v,1)

Solutions with zero prepended

Another numpy solution (just for completeness) is to use

numpy.ediff1d(v)

This works as numpy.diff, but only on a vector (it flattens the input array). It offers the ability to prepend or append numbers to the resulting vector. This is useful when handling accumulated fields that is often the case fluxes in meteorological variables (e.g. rain, latent heat etc), as you want a resulting list of the same length as the input variable, with the first entry untouched.

Then you would write

np.ediff1d(v,to_begin=v[0])

Of course, you can also do this with the np.diff command, in this case though you need to prepend zero to the series with the prepend keyword:

np.diff(v,prepend=0.0) 

All the above solutions return a vector that is the same length as the input.

My way

>>>v = [1,2,3,4,5]
>>>[v[i] - v[i-1] for i, value in enumerate(v[1:], 1)]
[1, 1, 1, 1]