CODEWITHSUNDEEP
pythonpython-3.x
Piotrek Wcisło
Piotrek Wcisło

Reputation: 85

Lists in python 3

So the problem is that I have a list which is made of pairs of numbers [ (0,0),(0,1)(0,2) ...etc and I would like to know how to delete from this list all pairs with the same numbers . List was created by this function. 

 l1 = []
 def fun2(x,y):
    for x in range(x+1):
        for y in range(y+1):
            l1.append((x,y))
    return l1

Upvotes: 0

Views: 86

Answers (2)

mhawke
mhawke

Reputation: 87064

You can avoid duplicate tuple elements while generating the list. Just add an if:

def fun2(x, y):
    result = []
    for a in range(x+1):
        for b in range(y+1):
            if a != b:
                result.append((a,b))
    return result

This could also be written more succinctly as a list comprehension:

result = [(a, b) for a in range(x+1) for b in range(y+1) if a != b]

Yet another option is to use itertools.product():

from itertools import product

result = [(a, b) for a, b in product(range(x+1), range(y+1)) if a != b]

Removing items afterwards is also possible with a list comprehension:

result = [pair for pair in result if pair[0] != pair[1]]

which creates a new list without the duplicate items then rebinds it to result. You can overwrite the list in place with:

result[:] = [pair for pair in result if pair[0] != pair[1]]

Upvotes: 5

Xero Smith
Xero Smith

Reputation: 2076

Method 1: Using list comprehnsion:

lst = [c for c in l1 if c[0] != c[1]]

Method 2: Building the list manually:

lst = []
for elem in l1:
    if elem[0] != elem[1]:
        lst.append(elem)

Upvotes: 0

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