如何划分Python 3中的列表列表？

Question

list1 = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0],
         [0, 500000.0, 500000.0, 500000.0], [0, 0, 1000000.0, 0],
         [0, 1000000.0, 500000.0, 2500000.0]]

list2 = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0],
         [0, 1, 1, 1], [0, 0, 2, 0], [0, 2, 1, 4]]

Can we divide each element from list1 and list2?

Output should again be in lists of list.

Answer 1

IIUC

import numpy as np
>>> np.array(list1)/list2
array([[    nan,     nan,     nan,     nan],
       [    nan,     nan,     nan,     nan],
       [    nan,     nan,     nan,     nan],
       [    nan,     nan,     nan,     nan],
       [    nan, 500000., 500000., 500000.],
       [    nan,     nan, 500000.,     nan],
       [    nan, 500000., 500000., 625000.]])

Answer 2

This is commonly known as zipwith . Python doesn't have a builtin function to do this, but it's easy to build yourself with a list comprehension.

[f(a, b) for a, b in zip(list1, list2)]  # where f is the function to zip with!

This is actually a zipwith of zipwith s, though, so let's nest:

[[aa/bb for (aa, bb) in zip(a, b)] for (a, b) in zip(list1, list2)]

---

EDIT: as Aran-Fey points out, zipwith does exist as map in Python, which makes this:

import functools
import operator

zipwithdiv = functools.partial(map, functools.partial(map, operator.truediv))

zipwithdiv(list1, list2)  # magic!

which is, admittedly, uglier than sin. But it makes my little functional heart go a-pitter patter.

Answer 3

You can try the following using the built-in function zip :

result = []
for sub_list1, sub_list2 in zip(list1, list2):
    sub_list = []
    for a, b in zip(sub_list1, sub_list2):
        if b == 0:
            sub_list.append('DivisionByZero')
            continue
        sub_list.append(a / b)
    result.append(sub_list)

this will produce:

[['DivisionByZero', 'DivisionByZero', 'DivisionByZero', 'DivisionByZero'],
 ['DivisionByZero', 'DivisionByZero', 'DivisionByZero', 'DivisionByZero'],
 ['DivisionByZero', 'DivisionByZero', 'DivisionByZero', 'DivisionByZero'],
 ['DivisionByZero', 'DivisionByZero', 'DivisionByZero', 'DivisionByZero'],
 ['DivisionByZero', 500000.0, 500000.0, 500000.0],
 ['DivisionByZero', 'DivisionByZero', 500000.0, 'DivisionByZero'],
 ['DivisionByZero', 500000.0, 500000.0, 625000.0]]