遍历字典列表和列表值

Question

Suppose I have a list of two dictionaries:

iter_list = [{0: [2, 1, 3], 
              1: [3, 2, 1], 
              2: [1, 2, 3], 
              3: [2, 3, 1], 
              4: [1, 2, 3], 
              5: [2, 3, 1]},
             {0: [2, 3, 1], 
              1: [1, 2, 3], 
              2: [1, 2, 3], 
              3: [1, 2, 3], 
              4: [2, 3, 1], 
              5: [1, 3, 2]}]

Each dictionary has 6 keys numbered 0 through 5. I would like to loop through each dictionary one at a time in the order of the key and have the output be a pair with the key (ordered by the key value) and the first value of the corresponding list, the first value corresponding with the second key etc. followed by the key and the second value etc. Hopefully the example output will clarify:

0 2
1 3
2 1
3 2
4 1
5 2
0 1
1 2
2 2
3 3
4 2
5 3
0 3
1 1
2 3
3 1
4 3
5 1
0 2 //2nd dictionary iteration 
1 1
2 1
3 1
4 2
5 1
0 3 
1 2
2 2
3 2
4 3
5 3
0 1
1 3
2 3
3 3
4 1
5 2

I've only been able to figure out how to loop through the first position of the first dictionary, but can't figure out how to use the [0] as an iteration variable to go through all positions in the value lists before moving on to the next dictionary.

for i in iter_list:
    for key, value in i.iteritems():
            print key, value[0]

Any help is greatly appreciated! Thanks!

Answer 1

If I understand you correctly, in your case, you need to iterate the iter_list 3 times

>>> for i in iter_list:
...     for x in range(3):
...         for key, value in i.iteritems():
...             print key, value[x]
...

Output:

0 2
1 3
2 1
3 2
4 1
5 2
0 1
1 2
2 2
3 3
4 2
5 3
0 3
1 1
2 3
3 1
4 3
5 1
0 2
1 1
2 1
3 1
4 2
5 1
0 3
1 2
2 2
3 2
4 3
5 3
0 1
1 3
2 3
3 3
4 1
5 2

Answer 2

A different approach using zip and itertools.chain and unpacking tricks:

from itertools import chain
packed_pairs = zip(*[[(k, v) for v in vs] for ds in iter_list for k, vs in ds.items()])
pairs = chain(*packed_pairs)
for pair in pairs:
    print pair[0], pair[1]

Output:

0 2
1 3
2 1
3 2
4 1
5 2
0 2
1 1
2 1
3 1
4 2
5 1
0 1
1 2
2 2
3 3
4 2
5 3
0 3
1 2
2 2
3 2
4 3
5 3
0 3
1 1
2 3
3 1
4 3
5 1
0 1
1 3
2 3
3 3
4 1
5 2

Note: make sure to use pairs = list(pairs) if you want to re-use pairs .

Answer 3

A list comprehension should do:

>>> output_list = [[k, d[k][i]] for d in iter_list for i in range(3) for k in sorted(d)]: 
>>> for k, dki in output_list:
...     print k, dki
...
0 2
1 3
2 1
3 2
4 1
5 2
0 1
1 2
2 2
3 3
4 2
5 3
0 3
1 1
2 3
3 1
4 3
5 1
0 2
1 1
2 1
3 1
4 2
5 1
0 3
1 2
2 2
3 2
4 3
5 3
0 1
1 3
2 3
3 3
4 1
5 2

Answer 4

perhaps you want to do this:

for i in iter_list:
    for j in range(3):
        for key, value in i.iteritems():
            print key, value[j]

assuming the values in the dictionaries have a fixed length of 3, otherwise you have to figure out where 3 comes from. For example, it could be the minimum value length across all values of all dictionaries:

>>> min(len(v) for k,v in i.iteritems() for i in iter_list)
3