Python: how to programmatically create a dictionary of arrays

Question

I have an array a that I want to store every 5 values of in a dictionary, while programmatically creating keys.

For example:

if we have

a=[1,2,3,4,5,6,7,8,9,10]

I want the dictionary to look like this:

d={"first_title":[1,2,3,4,5], "second_title":[6,7,8,9,10]}

Edit: I have thousands of terms so I want to do this in a loop

`a` is a list, not an array. Slicing into regular chunks is handled in many other questions and other help sites. Where are you stuck on that part? — Prune, Aug 17 '18 at 17:53
The `d` in your example is not a dictionary, it uses illegal notation. — DYZ, Aug 17 '18 at 18:02

blue_note · Answer 1 · 2018-08-17T18:29:34.957

3

Dictionary values can be any object, including arrays. So, you just start with an empty dict, and add array values to it.

my_dict = {}
year = 1984
for index in range(0, len(a), 5):
    my_dict[year] = a[index: index+5]
    year += 1

edited Aug 17 '18 at 18:29

answered Aug 17 '18 at 17:50

blue_note

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I have thousands of terms, so I wanted to do this in some type of loop – BobelLegend Aug 17 '18 at 17:56
@BobelLegend: what are the keys of the dictionary?? – blue_note Aug 17 '18 at 17:57
1

I want the keys to be a year. Such as 1984, 1985, etc. – BobelLegend Aug 17 '18 at 17:59
1

If you want the keys to be years, why don't you say so in the question? – DYZ Aug 17 '18 at 18:03
1

@Current revision of your answer: you wrote it under the wrong assumption that the array contains entries [1,2,...n] so you substituted it with `range(n)`. The op stated that he has an [arbitary] array so one could assume that the numbers are only an example... – Fabian N. Aug 17 '18 at 18:21
1

@FabianN.:edited – blue_note Aug 17 '18 at 18:29
I like this one, its clean, short, pythonic and works on an arbitrary length of a, would be really nice if we would now the actual use case... – Fabian N. Aug 17 '18 at 18:41

Dani Mesejo · Answer 2 · 2018-08-17T18:04:33.833

A somewhat general and pythonic solution is the following:

from itertools import izip_longest


def grouper(n, iterable, fillvalue=None):
    """grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"""
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

keys = ["first-title", "second-title"]
a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

result = {key: list(group) for key, group in zip(keys, grouper(5, a))}

This uses the grouper recipe (see this answer for a better explanation). A less pythonic solution, is to iterate over the pair of keys and groups using a for loop:

result = {}
for key, group in zip(keys, grouper(5, a)):
    result[key] = group

In both cases the output is:

{'first-title': [1, 2, 3, 4, 5], 'second-title': [6, 7, 8, 9, 10]}

Slightly offtopic but: +1 for using dict comprehension, I didn't even know that this is possible. — Fabian N., Aug 17 '18 at 20:17

Fabian N. · Accepted Answer · 2018-08-17T18:37:48.900

How about this?

Input:

a = [1,2,3,4,5,6,7,8,9,10]
keys = [1984, 1985, 1986, 1987] 
n = 5

Code:

my_dict = {}

for i in range(len(a)//n):
    key = str(keys[i]) if i < len(keys) else "after " + str(keys[-1])
    my_dict[key] = a[i*n: (i+1)*n]

print(my_dict)

Output:

{'1984': [1, 2, 3, 4, 5], '1985': [6, 7, 8, 9, 10]}

Depending on your use case you could also do something like this:

# Input
a = range(22)
keys = [1984, 1985, 1986] # maybe replace it with range(1984, 2000)
n = 5

# Code
b = a
my_dict = {}

for i in range(min(len(keys), len(a)//n)):
    key = keys[min(i, len(keys)-1)]
    my_dict[key] = b[:n]
    b = b[n:]

my_dict['later'] = b

print(my_dict)

# Output
{
    1984: [0, 1, 2, 3, 4], 
    1985: [5, 6, 7, 8, 9], 
    1986: [10, 11, 12, 13, 14], 
    'later': [15, 16, 17, 18, 19, 20, 21]
}

Python: how to programmatically create a dictionary of arrays

3 Answers3