From this
a = ['foo','bar','cat','dog']
to create another variable b that will have it as follows:
b = [['foo','bar'],['bar','cat'],['cat','dog']] # with this order
I have tried this:
b = [[i] * 2 for i in a]
but gives this:
[['foo', 'foo'], ['bar', 'bar'], ['cat', 'cat'], ['dog', 'dog']]
No external libraries please.
>Solution :
You can use a list comprehension like you attempted to, you just need a slight modification to account for different spots in the list:
b = [[a[i-1], a[i]] for i in range(1, len(a))]
Output:
[['foo', 'bar'], ['bar', 'cat'], ['cat', 'dog']]
Since you want to pair "touching" items the a[i-1] and a[i] accomplish this. Then just loop through each i in len(a). We use range(1,len(a)) instead of range(len(a)), however, because we don’t want to pair the first value of a with something before it because it is the first value.
An arguably cleaner solution would be to use zip, but it depends on preference:
b = [list(i) for i in zip(a, a[1:])]
In this case we use a[1:] in order to offset that list so that the pairs are correct.