如何从两个列表中创建一个列表列表

Question

I am trying to accomplish the following task and failing so far. I have two sorted lists of numbers. Say

A = [5.8, 6.5, 7.6, 14.5, 18.1, 25.7, 26.4, 30.7, 30.9, 33.6, 38.6, 38.8, 39.2]

B = [9.0, 13.5, 22.5, 32.3 40.6, 43.2, 47.9, 54.2, 60.3, 63.0]

I want to make a list of lists. Each list contains a different value from B and all the values in A between that B-value and the one directly before it in the list. The first list just contains everything smaller than the first value in B.

So in this case the list of lists should start

[[5.8, 6.5, 7.6, 9.0], [13.5], [14.5, 18.1, 22.5], [25.7, 26.4, 30.7. 30.9, 32.3]....]

I would be happy with numpy code if that is simpler/faster.

I tried:

[[*a, b] for b, a in itertools.groupby(A, lambda x: next(filter(lambda y: y >= x, B)))]

but it misses out all the singleton lists and I am not sure how fast it is in any case.

Answer 1

I would use two pointers like this,

i = 0
j= 0
ans = [] #contains the lists of lists
while j<len(B):
    to_append = []
    while i<len(A) and A[i]<=B[j]:
         to_append.append(A[i])
         i=i+1
    to_append.append(B[j])
    ans.append(to_append)
    j=j+1

Try now

Answer 2

You can use heapq.merge for an O(n+m) solution:

from itertools import chain, repeat
from heapq import merge

[*map(list, map(chain, map(iter, repeat(merge(A,B).__next__), B), zip(B)))]
# [[5.8, 6.5, 7.6, 9.0], [13.5], [14.5, 18.1, 22.5], [25.7, 26.4, 30.7, 30.9, 32.3], [33.6, 38.6, 38.8, 39.2, 40.6], [43.2], [47.9], [54.2], [60.3], [63.0]]

This uses merge to merge A and B in order. Next, the two argument form of iter is used to split at the elements from B. This unfortunately eats the split points, so we reattach them using itertools.chain.

Answer 3

you could do,

import numpy as np
A = [5.8, 6.5, 7.6, 14.5, 18.1, 25.7, 26.4, 30.7, 30.9, 33.6, 38.6, 38.8, 39.2]
B = [9.0, 13.5, 22.5, 32.3, 40.6, 43.2, 47.9, 54.2, 60.3, 63.0]
A = np.array(A)
B = np.hstack(([-np.inf], B))
result = [np.r_[np.extract((A>B[i]) & (A<= B[i+1]), A), B[i+1]] for i in range(len(B)-1)]

and if you want just lists, you could do afterwards :

result = [*map(lambda x : list(x), result),]
print(result)
# [[5.8, 6.5, 7.6, 9.0], [13.5], [14.5, 18.1, 22.5], [25.7, 26.4, 30.7, 30.9, 
# 32.3], [33.6, 38.6, 38.8, 39.2, 40.6], [43.2], [47.9], [54.2], [60.3], [63.0]]

Answer 4

Use np.searchsorted to solve the problem in O(N*logN) time. First, find position of elements from B in A. Next, split array A using those postions. Finally create a list of desired properties.

pos = np.searchsorted(A,B)
chunks = np.split(A, pos)
res = [np.hstack(ab) for ab in zip(chunks,B)]

This will generate the required list of ndarrays which you can convert back to list using ndarray.tolist() method:

res_list = list(map(np.ndarray.tolist, res))

Answer 5

As a purely Numpy-based approach (not that significant tho) you could convert your lists to array and do the following:

In [43]: ind = (b[:,None] > a).sum(1)

In [44]: np.split(np.insert(a, ind, b), ind + np.arange(1, ind.size +1))
Out[44]: 
[array([5.8, 6.5, 7.6, 9. ]),
 array([13.5]),
 array([14.5, 18.1, 22.5]),
 array([25.7, 26.4, 30.7, 30.9, 32.3]),
 array([33.6, 38.6, 38.8, 39.2, 40.6]),
 array([43.2]),
 array([47.9]),
 array([54.2]),
 array([60.3]),
 array([63.]),
 array([], dtype=float64)]