从排序列表中获取大于给定数字的第一个元素

Question

I have two lists. List B is like a database to which I need to compare each element of list A, one by one. Lets say

B = [0.6, 1.7, 3, 4.5]
A = [0.6, 0.9, 1.2, 1.5, 2, 2.5, 3, 4, 4.5]

B is a sorted list, so for each A[i] whenever the algorithm finds a number which is >= A[i] in B, it should return that as the output. So my output should look something like:

C = [0.6, 1.7, 1.7, 1.7, 3, 3, 3, 4.5, 4.5]

Could you please suggest me the simplest solution, avoiding nested loops as much as possible?

Answer 1

If you can use a 3rd party library, one solution is NumPy via np.searchsorted :

import numpy as np

B = np.array([0.6, 1.7, 3, 4.5])
A = [0.6, 0.9, 1.2, 1.5, 2, 2.5, 3, 4, 4.5]

res = B[np.searchsorted(B, A)]

array([ 0.6,  1.7,  1.7,  1.7,  3. ,  3. ,  3. ,  4.5,  4.5])

This will be more efficient than a sequential loop or an algorithm based on bisect from the standard library.

Answer 2

Just a next would do ( if I understood you correctly ):

A = [0.6, 0.9, 1.2, 1.5, 2, 2.5, 3, 4, 4.5]
B = [0.6, 1.7, 3, 4.5]

C = [next(b for b in B if b >= a) for a in A]

print(C)  # -> [0.6, 1.7, 1.7, 1.7, 3, 3, 3, 4.5, 4.5]

Answer 3

Since B is sorted, you can use bisect to binary-search the correct value in B :

>>> B = [0.6, 1.7, 3, 4.5]
>>> A = [0.6, 0.9, 1.2, 1.5, 2, 2.5, 3, 4, 4.5]
>>> import bisect
>>> [B[bisect.bisect_left(B, a)] for a in A]
[0.6, 1.7, 1.7, 1.7, 3, 3, 3, 4.5, 4.5]

This has complexity O(alogb) , with a and b being then lengths of A and B respectively. Assuming that A also is sorted, as in your example, you could also do it in O(a+b) :

i, C = 0, []
for a in A:
    while B[i] < a:
        i += 1
    C.append(B[i])

Note, however, that both approaches (as well as the other answers posted so far) will fail if A contains a number larger than any number in B .

Answer 4

Since your given B list is sorted, you could use:

B = [0.6, 1.7, 3, 4.5]
A = [0.6, 0.9, 1.2, 1.5, 2, 2.5, 3, 4, 4.5]

def first_greater_elem(lst, elem):
    for item in lst:
       if item >= elem:
         return item

Then just use a list comprehension .

C = [first_greater_elem(B,item) for item in A ]

Output

[0.6, 1.7, 1.7, 1.7, 3, 3, 3, 4.5, 4.5]

Another approach could be using bisect_left method from bisect package.

C = [B[bisect_left(B,item)] for item in A ]

Output

[0.6, 1.7, 1.7, 1.7, 3, 3, 3, 4.5, 4.5]

Answer 5

A recursive way (destructive for original lists), works also if list_a contains number larger than list_b :

def pick(lst, ref, res=None):
  if res == None: res = []
  if len(lst) == 0: return res
  if ref[0] >= lst[0]:
    res.append(ref[0])
    lst.pop(0)
  elif len(ref) == 1 and ref[0] < lst[0]:
    # res.extend(lst) # if want to append the rest of lst instead of stop the loop
    # or do whathever is best for you
    return res
  else: ref.pop(0)
  pick(lst, ref, res)
  return res


list_b = [0.6, 1.7, 3, 3.9]
list_bb = [0.5]
list_a = [0.6, 0.9, 1.2, 1.5, 2, 2.5, 3, 4, 4.5]

print(pick(list_a, list_b))
#=> [0.6, 1.7, 1.7, 1.7, 3, 3, 3]

print(pick(list_a, list_bb))
#=> []