python dictionary put values of keys into a list where the keys are in another list

Question

I have a dictionary which looks like

a = {32: [2230], 37265: [2218], 51: [2223], 164: [2227], 3944: [2224]}

however, values in a could contain multiple elements, like

a = {32: [2200, 2230], 37265: [2201, 2218], 51: [2223], 164: [2227], 3944: [2224]}

I have a list that stores the keys in a in groups,

b = [[32, 51, 164], [3944, 37265]]

now I want to get values of keys in each group in another list,

        clusters = []
        for key_group in b:
            group = []

            for key in key_group:
                group.extend(a[key])

            if len(group) > 1:
                clusters.append(group) 

so the final list looks like,

clusters = [[2230, 2223, 2227], [2224, 2218]]

if a contains multiple elements in a value list, clusters looks like,

clusters = [[2200, 2230, 2223, 2227], [2224, 2201, 2218]]

I am wondering what's the best way to do this.

Also in case when b contains list(s) which has only one value/key, and if this key maps to a single element list in a , this list will be ignored,

a = {32: [2200, 2230], 37265: [2201, 2218], 51: [2223], 164: [2227], 3944: [2224]}

b = [[32, 51, 164], [3944], [37265]]

while 3944 maps to [2224] , which will be ignored, but 37265 maps to [2201, 2218] , which will be kept, since len([2201, 2218]) > 1 . clusters will look like the following in this case,

clusters = [[2200, 2230, 2223, 2227], [2201, 2218]]         

Answer 1

Assuming your values in a are always just a list with a single element, then it is a simple nested list comprehension :

[[a[k][0] for k in sublist] for sublist in b]
# [[2230, 2223, 2227], [2224, 2218]]

Since you've now clarified the values of a can be lists with multiple elements, you can use itertools.chain.from_iterable to flatten the lists returned and give your desired output:

from itertools import chain
[list(chain.from_iterable(a[k] for k in sublist)) for sublist in b]
# [[2200, 2230, 2223, 2227], [2224, 2201, 2218]]

Answer 2

You can use list comp to do the same job:

clusters = [[a[key][0] for key in k] for k in b]

res => [[2230, 2223, 2227], [2224, 2218]]

if a could have multiple elems, you can do also:

clusters = [[it for key in k for it in a[key]] for k in b]

res => [[2200, 2230, 2223, 2227], [2224, 2201, 2218]]

Answer 3

I would suggest using a triple list comprehension:

list(filter(lambda l:len(l) > 1, ([elem for key in lst for elem in a[key]] for lst in b)))

This works for list values of arbitrary length. filter removes the lists with one or zero elements.

Answer 4

根据你的编辑：

[[item for i in sub for item in a[i]] for sub in b]