How to compare each item in a list with the rest only once

Question

Say I have an array list of things I want to compare  In languages I am more familiar with  I would do something like  for  int i   0  i  lt  mylist size    i        for  int j   i   1  j  lt  mylist size    j            compare mylist i   mylist j     This ensures we only compare each pair once  For some context  I am doing collision detection on a bunch of objects contained in the list  For each collision detected  a small  collision  object describing the collision is appended to a list  which another routine then loops through resolving each collision  depending on the nature of the two colliding objects   Obviously  I only want to report each collision once   Now  what is the pythonic way of doing this  since Python favors using iterators rather than looping over indices   I had the following  buggy  code   for this in mylist      for that in mylist          compare this  that    But this clearly picks up each collision twice  which lead to some strange behavior when trying to resolve them  So what is the pythonic solution here

User · Answer

This code will count frequency and remove duplicate elements   from collections import Counter  str1  the cat sat on the hat hat   int list str1 split     unique list      for el in int list       if el not in unique list          unique list append el      else          print  Element already in the list   print unique list  c Counter int list   c values    c keys    print c

User · Answer

Use itertools combinations mylist  2   mylist   range 5  for x y in itertools combinations mylist  2       print x y  0 1 0 2 0 3 0 4 1 2 1 3 1 4 2 3 2 4 3 4

User · Answer

Your solution is correct  but your outer loop is still longer than needed  You don t need to compare the last element with anything else because it s been already compared with all the others in the previous iterations  Your inner loop still prevents that  but since we re talking about collision detection you can save the unnecessary check   Using the same language you used to illustrate your algorithm  you d come with something like this   for  int i   0  i  lt  mylist size   - 1    i      for  int j   i   1  j  lt  mylist size    --j          compare mylist i   mylist j

User · Answer

I think using enumerate on the outer loop and using the index to slice the list on the inner loop is pretty Pythonic   for index  this in enumerate mylist       for that in mylist index 1            compare this  that

User · Answer

Of course this will generate each pair twice as each for loop will go through every item of the list   You could use some itertools magic here to generate all possible combinations   import itertools for a  b in itertools combinations mylist  2       compare a  b    itertools combinations will pair each element with each other element in the iterable  but only once     You could still write this using index-based item access  equivalent to what you are used to  using nested for loops   for i in range len mylist        for j in range i   1  len mylist            compare mylist i   mylist j     Of course this may not look as nice and pythonic but sometimes this is still the easiest and most comprehensible solution  so you should not shy away from solving problems like that

[python] How to compare each item in a list with the rest, only once?

Examples related to python