根据另一个列表 python 计算列表中所有唯一元素的总和

Question

I have two lists like this,

a=[['a', 'b', 'c'], ['b', 'c'], ['a', 'd'], ['x']]
b=[[1, 2, 3], [4,5], [6,7], [8]] (the size of a and b is always same)

Now I want to create two list with the sum of unique elements, so the final lists should look like,

 a=['a', 'b', 'c', 'd', 'x']
 b=[7, 6, 8, 7, 8] (sum of all a, b, d, d and x)

I could do this using for loop but looking for some efficient way to reduce execution time.

Answer 1

Not so pythonic but will do the job:

a=[['a', 'b', 'c'], ['b', 'c'], ['a', 'd'], ['x']]
b=[[1, 2, 3], [4,5], [6,7], [8]]

mapn = dict()
for elt1, elt2 in zip(a, b):
    for e1, e2 in zip(elt1, elt2):
        mapn[e1] = mapn.get(e1, 0) + e2

elts = mapn.keys()
counts = mapn.values()

print(mapn)
print(elts)
print(counts)

Answer 2

You can use zip and collections.Counter along the following lines:

from collections import Counter

c = Counter()
for la, lb in zip(a, b):
    for xa, xb in zip(la, lb):
        c[xa] += xb
 
list(c.keys())
# ['a', 'b', 'c', 'd', 'x']
list(c.values())
# [7, 6, 8, 7, 8]

Answer 3

Here some ideas.

First, to flatten your list you can try:

a=[['a', 'b', 'c'], ['b', 'c'], ['a', 'd'], ['x']]
b=[[1, 2, 3], [4,5], [6,7], [8]]

To have uniques elements, you can do something like

A = set([item for sublist in a for item in sublist])

But what I would do first (perhaps not the more efficient) is:

import pandas as pd
import bumpy as np
LIST1 = [item for sublist in a for item in sublist]
LIST2 = [item for sublist in b for item in sublist]
df = pd.DataFrame({'a':LIST1,'b':LIST2})
df.groupby(df.a).sum()

OUTPUT:

Answer 4

At the end of the day, you're going to have to use two for loops. I have a one liner solution using zip and Counter .

The first solutions works only in this specific case where all the strings are a single character, because it creates a string with the right number of each letter, and then gets the frequency of each letter.

from collections import Counter

a = [['a', 'b', 'c'], ['b', 'c'], ['a', 'd'], ['x']]
b = [[1, 2, 3], [4,5], [6,7], [8]]

a, b = zip(*Counter(''.join(x*y for al, bl in zip(a, b) for x, y in zip(al, bl))).items())

For the more general case, you can do:

a, b = zip(*Counter(dict(p for al, bl in zip(a, b) for p in zip(al, bl))).items())

Answer 5

You can combine the lists and their internal lists using zip(), then feed the list of tuples to a dictionary constructor to get a list of dictionaries with values for each letter. Then convert those dictionaries to Counter and add them up.

a = [['a', 'b', 'c'], ['b', 'c'], ['a', 'd'], ['x']]
b = [[ 1,   2,   3 ], [ 4,   5 ], [ 6,   7 ], [ 8 ]]

from collections import Counter
from itertools import starmap

mapn        = sum(map(Counter,map(dict,starmap(zip,zip(a,b)))),Counter())
elts,counts = map(list,zip(*mapn.items()))

print(mapn)   # Counter({'c': 8, 'x': 8, 'a': 7, 'd': 7, 'b': 6})    
print(elts)   # ['a', 'b', 'c', 'd', 'x']    
print(counts) # [ 7,   6,   8,   7,   8]

detailed explanation:

• zip(a,b) combines the lists into pairs of sublists. eg (['a','b','c'],[1,2,3]), ...
• starmap(zip,...) takes these list pairs and merges then together into sublist of letter-number pairs: [('a',1),('b',2),('c',3)], ...
• Each of these lists of pairs is converted to a dictionary by map(dict,...) and then into a Counter object by map(counter,...)
• We end up with a list of Counter objects corresponding to the pairing of each sublist. Applying sum(...,Counter()) computes the totals for each letter into a single Counter object.
• Apart from being a Counter object, mapn is excatly the same as the dictionary that you produced.