Python Finding differences between elements of a list

Question

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

User · Answer

Using the    walrus operator available in Python 3 8     gt  gt  gt  t    1  3  6   gt  gt  gt  prev   t 0    -prev    prev    x  for x in t 1     2  3

User · Answer

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

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If you don t want to use numpy nor zip  you can use the following solution    gt  gt  gt  t    1  3  6   gt  gt  gt  v    t i 1 -t i  for i in range len t -1    gt  gt  gt  v  2  3

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gt  gt  gt  t  1  3  6   gt  gt  gt   j-i for i  j in zip t  -1   t 1        or use itertools izip in py2k  2  3

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I suspect this is what the numpy diff command does anyway  but just for completeness you can simply difference the sub-vectors  from numpy import array as a a x 1   -a x  -1    In addition  I wanted to add these solutions to generalizations of the question  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

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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

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Ok  I think I found the proper solution  v    x 0 -x 1  for x in zip t 1   t  -1

User · Answer

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

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My way   gt  gt  gt v    1 2 3 4 5   gt  gt  gt  v i  - v i-1  for i  value in enumerate v 1    1    1  1  1  1

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A functional approach    gt  gt  gt  import operator  gt  gt  gt  a    1 3 5 7 11 13 17 21   gt  gt  gt  map operator sub  a 1    a  -1    2  2  2  4  2  4  4    Using generator    gt  gt  gt  import operator  itertools  gt  gt  gt  g1 g2   itertools tee  x x for x in xrange 5   2   gt  gt  gt  list itertools imap operator sub  itertools islice g1 1 None   g2    1  3  5  7    Using indices    gt  gt  gt   a i 1 -a i  for i in xrange len a -1    2  2  2  4  2  4  4

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In the upcoming Python 3 10 release schedule  with the new pairwise function it s possible to slide through pairs of elements and thus map on rolling pairs  from itertools import pairwise   y-x for  x  y  in pairwise  1  3  6  7       2  3  1    The intermediate result being  pairwise  1  3  6  7       1  3    3  6    6  7

[python] Python: Finding differences between elements of a list

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