Enumerate list of elements starting from the second element

Question

I have this list

[[\'a\', \'a\', \'a\', \'a\'],
 [\'b\', \'b\', \'b\', \'b\', \'b\'],
 [\'c\', \'c\', \'c\', \'c\', \'c\']]

and I want to conca

Answer 1

You can explicitly create an iterable with the iter() builtin, then call `next(iterable) to consume one item. Final result is something like this:

line_iter = iter(list_of_lines[:])
# consume first item from iterable
next(line_iter)
for index, item in enumerate(line_iter, start=1):
    list_of_lines[index][1:3] = [''.join(item[1:3])]

Note the slice on the first line, in general it's a bad idea to mutate the thing you're iterating over, so the slice just clones the list before constructing the iterator, so the original list_of_lines can be safely mutated.

Answer 2

When you call

enumerate(list_of_lines, start=1)

, the pairs that it generates are not

1 ['b', 'b', 'b', 'b', 'b']
2 ['c', 'c', 'c', 'c', 'c']

, but rather

1 ['a', 'a', 'a', 'a']
2 ['b', 'b', 'b', 'b', 'b']
3 ['c', 'c', 'c', 'c', 'c']

That is, the start value indicates what the first index used should be, not what the first element to use is.

Perhaps an alternate way of doing this would be as follows:

for (index, item) in list(enumerate(list_of_lines))[1:]:
    list_of_lines[index][1:3] = [''.join(item[1:3])]

Answer 3

There is not much merit here for using enumerate() ... you can simply .pop() the n-th item from inner lists. For loop over your data, starting at index 1 and add the 2nd value (popped) to the 1st element of the inner list:

data = [['a', 'a', 'a', 'a'],
 ['b', 'b', 'b', 'b', 'b'],
 ['c', 'c', 'c', 'c', 'c']] 

for row in range(1,len(data)):  # apply to 1st to n-th inner list by index
    item = data[row].pop(2)         # remove the 2nd item from inner list
    data[row][1] += item            # add it to the 1st of inner list

print(data)

Output:

[['a', 'a', 'a', 'a'], 
 ['b', 'bb', 'b', 'b'], 
 ['c', 'cc', 'c', 'c']]

See list.pop(index)