在新行上为字典中的每个键输出多个值，并在键之前

Question

In Python, I have two lists, ids and gos and both have the same length (ie equal number of lines), but gos can have multiple elements per line (ie, it is a list of lists) while ids has only 1 element per line

For example:

ids =  ['1','2','3']
gos =  [['a','b','c'],['d', 'e'], ['f']]

I want to print out each id in ids as many times as there are go-elements in the gos list followed by one of the corresponding elements from gos list, and each time on a new line.

I hope this clarifies the output I am seeking:

'1'   'a'
'1'   'b'
'1'   'c'
'2'   'd'
'2'   'e'
'3'   'f'

Answer 1

Use zip:

for i,g in zip(ids, gos):
   for ge in g:
      print i,ge

output:

1 a
1 b
1 c
2 d
2 e
3 f

Answer 2

You can try these simple nested for loops:

for i, e1 in enumerate(ids):
    for e2 in gos[i]:
        print e1, e2

Output:

1 a
1 b
1 c
2 d
2 e
3 f

Note: The enumerate function is used to keep count of the index, which will be used to access to the corresponding element of gos .

Answer 3

another way of doing this

from itertools import product,imap
ids =  ['1','2','3']
gos =  [['a','b','c'],['d', 'e'], ['f']]

for i in imap(product,ids,gos):
    for j in i:
        print j[0],j[1]

Answer 4

Yet another way of doing it..

ids = "1 2 3".split()
gos = ['a b c'.split(), 'c d'.split(), ['f']]

import itertools

ig = [zip([i] * len(g), g)
      for i, g in zip(ids, gos)]

for i, g in itertools.chain.from_iterable(ig):
    print(i, g)

Answer 5

I hope this is an easy way to do that

for i in range(len(gos)):
    for k in gos[i]:
        print ids[i], k

Output

1 a
1 b
1 c
2 d
2 e
3 f