将一个列表的每个元素与循环中的第二个列表的四个元素相乘以获取新列表的函数

Question

我有两个列表，一个长度为40，另一个长度为10.我想将40个列表中的前四个元素与第二个列表的第一个元素相乘，并将整个第一个列表中的40个循环为一个新的列表是这两者的产物。关于如何解决这个问题的任何建议？

from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

v=[[a1/ b1 for a1 in a4] for a4, b1 in zip(grouper(active, 4), passive)]

active=[56.93977737426758,
 54.12062072753906,
 54.89398765563965,
 55.214101791381836,
 54.29464149475098,
 53.80845832824707,
 54.46353721618652,
 54.49761962890625,
 53.01671028137207,
 53.872962951660156,
 53.156455993652344,
 53.20746994018555,
 52.529762268066406,
 56.03120040893555,
 54.122426986694336,
 55.83853149414063,
 53.51207160949707,
 54.82537269592285,
 53.569284439086914,
 53.5296745300293,
 54.354637145996094,
 54.313310623168945,
 53.26720809936523,
 54.64541053771973,
 55.00912475585938,
 55.093666076660156,
 55.138763427734375,
 54.19987297058106,
 54.07197189331055,
 53.18226623535156,
 53.656246185302734,
 54.97188377380371,
 55.28757095336914,
 54.08882141113281,
 53.08153915405274,
 53.61944770812988,
 53.15986633300781,
 53.53702735900879,
 53.32623863220215,
 52.01462173461914]

passive= [54.46392059326172,
 52.37292861938477,
 51.95756149291992,
 53.40110778808594,
 54.46831512451172,
 56.04657173156738,
 57.74487495422363,
 53.75052452087402,
 56.246402740478516,
 55.15713691711426]

我当前的输出是一个10.的列表。我想要一个40的列表。我想把活动的前四个元素除以被动的第一个元素...依此类推。最后我想要一个包含40个元素的新列表而不是10.例子[active1 / passive1，active2 / passive1，active3 / passive1 .... active40 / passive10]

[[1.0454586587604597,
  0.9936967470945319,
  1.0078963662125922,
  1.0137739110579735],
 [1.0366928664488493,
  1.0274097658218595,
  1.0399177333006342,
  1.040568497228071],
 [1.020384882546816,
  1.0368647296698332,
  1.0230744951511204,
  1.0240563338877238],
 [0.9836830066620091,
  1.0492516490722763,
  1.013507569945381,
  1.045643691807429],
 [0.9824440408551517,
  1.0065553261670555,
  0.9834944282126284,
  0.982767218109524],
 [0.9698119879004422,
  0.9690746274955083,
  0.9504097477092123,
  0.9750000553011796],
 [0.9526234977470646,
  0.954087546649548,
  0.9548685224696527,
  0.9386092361191739],
 [1.005980357871888,
  0.9894278559960512,
  0.9982460015709302,
  1.0227227411047006],
 [0.9829530113857514,
  0.9616405454531767,
  0.9437321600631335,
  0.9532955903958973],
 [0.9637894441999811,
  0.9706273811757119,
  0.9668057773255447,
  0.9430261366318299]]

Answer 1

对于例如

a = range(40)
b = range(10)

最简单的：

[x * b[i//4] for i, x in enumerate(a)]

更多功能：

# from https://docs.python.org/3/library/itertools.html#itertools-recipes
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

[a1 * b1 for a4, b1 in zip(grouper(a, 4), b) for a1 in a4]

Answer 2

你可以在一行中使用纯Python来做到这一点：

newList = [active[j] / passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
for elem in newList:
    print(elem)

这将打印：

1.0454586587604597
0.9936967470945319
1.0078963662125922
1.0137739110579735
1.0366928664488493
1.0274097658218595
1.0399177333006342
1.040568497228071
1.020384882546816
1.0368647296698332
1.0230744951511204
1.0240563338877238
0.9836830066620091
1.0492516490722763
1.013507569945381
1.045643691807429
0.9824440408551517
1.0065553261670555
0.9834944282126284
0.982767218109524
0.9698119879004422
0.9690746274955083
0.9504097477092123
0.9750000553011796
0.9526234977470646
0.954087546649548
0.9548685224696527
0.9386092361191739
1.005980357871888
0.9894278559960512
0.9982460015709302
1.0227227411047006
0.9829530113857514
0.9616405454531767
0.9437321600631335
0.9532955903958973
0.9637894441999811
0.9706273811757119
0.9668057773255447
0.9430261366318299

乘法也是如此：

newList = [active[j] * passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]