Compute List Difference

Compute list difference

If the order does not matter, you can simply calculate the set difference:

>>> set([1,2,3,4]) - set([2,5])
set([1, 4, 3])
>>> set([2,5]) - set([1,2,3,4])
set([5])

Get difference between two lists

To get elements which are in temp1 but not in temp2 :

In [5]: list(set(temp1) - set(temp2))
Out[5]: ['Four', 'Three']

Beware that it is asymmetric :

In [5]: set([1, 2]) - set([2, 3])
Out[5]: set([1]) 

where you might expect/want it to equal set([1, 3]). If you do want set([1, 3]) as your answer, you can use set([1, 2]).symmetric_difference(set([2, 3])).

Python: Finding differences between elements of a list

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

Compute the differences between the elements of two lists in python

Problem

Most Pythonic way to efficiently calculate the cartesian product of iterables and subtract the results.

Solution

The simplest and most pythonic approach:

[ abs(i-j) for i,j in itertools.product(list1, list2) ]

Depending on how large your dataset is, you may need to use a generator expression instead.

( abs(i-j) for i,j in itertools.product(list1, list2) )

Note that you don't need to explicitly define l1 and l2 since itertools.product takes n iterables. You can simply do:

l = [[1], [3,4]]

( abs(i-j) for i,j in itertools.product(*l) )

References

itertools.product: https://docs.python.org/3/library/itertools.html#itertools.product

generator expressions: https://www.python.org/dev/peps/pep-0289/

Trying to compute difference in a list from the first value of the list

First, a list in python is written with square brackets, like that:

l = [1, 3, 5, 5, 2, 4, 6]

To get your result there are many ways. One way would be list comprehensions, for example with

 [x - l[0] for x in l]

You can use map:

map(lambda x: x-l[0], l)

You could use a loop of many other ways.

Calculate difference between adjacent items in a python list

The straight approach here is the best:

x = s[1] - s[0]
for i in range(2, len(s)):
    if s[i] - s[i-1] != x: break
else:
    #do some work here...

python find difference between two lists

You can convert the lists to sets and run the usual set operations such as difference or symmetric difference. For example, set(b) - set(a) evaluates to set([7, 8, 9]).

Python, compute array difference in exact amount of elements

from collections import Counter

temp1 = ['A', 'A', 'A', 'B', 'C', 'C', 'C']
temp2 = ['A', 'B', 'C', 'C']

result = []

counts = Counter(temp2)

for item in temp1:
    if item in counts and counts[item]:
        counts[item] -= 1
    else:
        result.append(item)

print(result)

Output:

['A', 'A', 'C']

Scales O(n) and does not rely on sorted input.

This answer relies on the fact that Counter is just a subclass of dict, so we can use the instance as a mutable object in which to store the number of occurrences in temp2 that we still need to exclude from the result during the iteration over temp1. The documentation states explicitly that "Counter is a dict subclass" and that "Counter objects have a dictionary interface", which is a pretty good guarantee that item assignment will be supported, and that it is not necessary to treat it as a read-only object that must first be copied into a plain dict.

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