List of Dicts To/From Dict of Lists

List of dicts to/from dict of lists

Perhaps consider using numpy:

import numpy as np

arr = np.array([(0, 2), (1, 3)], dtype=[('a', int), ('b', int)])
print(arr)
# [(0, 2) (1, 3)]

Here we access columns indexed by names, e.g. 'a', or 'b' (sort of like DL):

print(arr['a'])
# [0 1]

Here we access rows by integer index (sort of like LD):

print(arr[0])
# (0, 2)

Each value in the row can be accessed by column name (sort of like LD):

print(arr[0]['b'])
# 2

List of dicts to/from dict of lists, but with all possible permutations

Make sure every value is a list and then use itertools.product:

from itertools import product

# DL = {'a': [0, 1], 'b': [2, 3]}
D = {'a': [0, 1], 'b': [2, 3], 'c': 7, 'd': 9}

res = [dict(zip(D, p)) for p in product(*[v if isinstance(v, list) else [v] for v in D.values()])]
print(res)

Output

[{'a': 0, 'b': 2, 'c': 7, 'd': 9}, {'a': 0, 'b': 3, 'c': 7, 'd': 9}, {'a': 1, 'b': 2, 'c': 7, 'd': 9}, {'a': 1, 'b': 3, 'c': 7, 'd': 9}]

Converting a dictionary with lists into a list of dictionaries

Using zip:

[dict(zip(d, vals)) for vals in zip(*d.values())]

Result:

[{'n': 'a', 'a': 1, 'p': '123'}, {'n': 'b', 'a': 2, 'p': '321'}, {'n': 'x', 'a': 3, 'p': '456'}]

How do I convert a list of dictionaries to a dictionary of lists in Python?

import collections

result = collections.defaultdict(list)

for d in dictionaries:
    for k, v in d.items():
        result[k].append(v)

Most pythonic way to convert a dict of lists into a list of dicts of all combs of the elements in the lists from the dict?

You could use itertools.product on just the values of the dict and then dict(zip(...)) the keys and values together to output a list of dicts.

from itertools import product

d = {'Apple':[3,5,7],'Pear':[5,2],'Orange':[10],'Banana':[7,10],'Lemon':[5,7,9,2,1]}

l = [dict(zip(d.keys(), vals)) for vals in product(*d.values())]
print(l)
# [{'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 1}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 1}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 1}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 1}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 1}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 1}]

Split dictionary of lists into list of dictionaries

Works for any number of keys

>>> map(dict, zip(*[[(k, v) for v in value] for k, value in d.items()]))
[{'key2': 4, 'key1': 1}, {'key2': 5, 'key1': 2}, {'key2': 6, 'key1': 3}]

For example:

d = {'key3': [7, 8, 9], 'key2': [4, 5, 6], 'key1': [1, 2, 3]}

>>> map(dict, zip(*[[(k, v) for v in value] for k, value in d.items()]))
[{'key3': 7, 'key2': 4, 'key1': 1}, {'key3': 8, 'key2': 5, 'key1': 2}, {'key3': 9, 'key2': 6, 'key1': 3}]

A general solution that works on any number of values or keys: (python2.6)

>>> from itertools import izip_longest
>>> d = {'key2': [3, 4, 5, 6], 'key1': [1, 2]}
>>> map(lambda a: dict(filter(None, a)), izip_longest(*[[(k, v) for v in value] for k, value in d.items()]))
[{'key2': 3, 'key1': 1}, {'key2': 4, 'key1': 2}, {'key2': 5}, {'key2': 6}]

And if you don't have python2.6:

>>> d = {'key2': [3, 4, 5, 6], 'key1': [1, 2]}
>>> map(lambda a: dict(filter(None, a)), map(None, *[[(k, v) for v in value] for k, value in d.items()]))
[{'key2': 3, 'key1': 1}, {'key2': 4, 'key1': 2}, {'key2': 5}, {'key2': 6}]

How to convert list of dictionaries into list of lists

I find this site that use blew code:
How to convert a list of dictionaries to a list of lists in Python

[[row[key] for row in myList] for key in keylist]

Test:

def bycol_decl(lod, keylist):
    return  [[row[key] for row in lod] for key in keylist]

if __name__ == "__main__": 
    lod = [
        {'a':1, 'b':2, 'c':3},
        {'a':4, 'b':5, 'c':6},
        {'a':7, 'b':8, 'c':9},
        ]
    keylist = ['a', 'b', 'c']
    print bycol_decl(lod, keylist)

Results:

[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

Convert a list of lists to a dictionary

You have zip as an option:

wanted = {a[0]: list(a[1:]) for a in zip(*x)}

Or if you're familiar with unpacking:

wanted = {k: v for k, *v in zip(*x)}