I have the following list
b = [[1,2], [1,3], [2,4], [2,5], [3,4], [3,5], [4,6], [5,6]]
I want to check the condition when second element of list b is equal to its first element.
len_b = len(b);
for i in range(0, len_b):
if b[i][1] == b[0][i]:
print 'anything'
but, whenever I execute this I have the IndexError: list index out of range.
Hackaholic's solutions fixes the issue but the Pythonic way to do this is to use tuple unpacking not indexing:-
>>> b = [[1,2], [1,3], [2,4], [2,5], [3,4], [3,5], [4,6], [5,6]]
>>> for x, y in b:
if x == y:
print 'anything'
In addition to the provided answers, there is a simple comprehension you can use to get a list of all items satisfying the criteria:
b = [[1,2], [1,3], [2,4], [2,5], [3,4], [3,5], [4,6], [5,6]]
c = [[x,y] for x, y in b if x == y]
This can be handy if you want to filter a list.
In your code , it should be:
b = [[1,2], [1,3], [2,4], [2,5], [3,4], [3,5], [4,6], [5,6]]
len_b = len(b) # not need of semi-colon here
for i in range(0, len_b): # range(len_b) is enough
if b[i][1] == b[i][0]: # see here
print 'anything'
Let me introduce you with filter and lambda:
filter(lambda x:x[0]==x[1], b)
Fix as:
len_b = len(b);
for i in range(0, len_b):
if b[i][0] == b[i][1]:
print 'anything'
Sensory, you could have more pythonic way to write code, like:
for x, y in b:
if x == y:
print 'xxx'
b[0][i]should beb[i][0]