2個列表的Python組合

Question

什麼是pythonic計算兩個列表的所有產品組合的方法。因此，給出兩個長度爲n的列表，我希望返回包含產品的長度爲2^n的列表。

像list(itertools.product(on,off))但結果應該使用所有四個要素，不僅組合對像：

[(1.05, 5.53), (1.05, 3.12), (1.05, 3.75), (1.05, 4.75), (1.5, 5.53), (1.5, 3.12), (1.5, 3.75), (1.5, 4.75), (2.1, 5.53), (2.1, 3.12), (2.1, 3.75), (2.1, 4.75), (1.7, 5.53), (1.7, 3.12), (1.7, 3.75), (1.7, 4.75)] 

因此，更多的是這樣的：

off = [5.53,3.12,3.75,4.75] 
on = [1.05,1.5,2.1,1.7] 

# calculate combinations 
x = combinations(on,off) 

# Where... 
# x[0] = off[0] * off[1] * off[2] * off[3] i.e 
# x[0] = 5.53 * 3.12 * 3.75 * 4.75 
# 
# x[1] = off[0] * off[1] * off[2] * on[3] i.e 
# x[1] = 5.53 * 3.12 * 3.75 * 1.7 
# 
# x[2] = off[0] * off[1] * on[2] * on[3] i.e 
# x[2] = 5.53 * 3.12 * 2.1 * 1.7 
# 
# ... 
# 
# x[15] = on[0] * on[1] * on[2] * on[3] i.e 
# x[15] = 1.05 * 1.5 * 2.1 * 1.7 

輸出可以是類似itertools.product()方法即[(5.53, 3.12, 3.75, 4.75),(5.53, 3.12, 3.75, 1.7), ...]我需要計算產品，但我在組合方法中很有趣。

注：當我說這樣做的pythonic方式我的意思是簡單的一兩行取蟒蛇結構的優勢，圖書館（itertools ECT）。

Answer 1

您可以通過itertools.product([0, 1], 4)開始生成所有2**4可能性。此列舉的0一切都可能序列和1小號長度爲4的，然後就可以通過利用off[i]如果0-1序列的第i個元素是0翻譯每0-1序列值從off和on，一個序列，否則爲on[i]。在代碼：

>>> import itertools 
>>> off = [5.53,3.12,3.75,4.75] 
>>> on = [1.05,1.5,2.1,1.7] 
>>> for choices in itertools.product([0, 1], repeat=len(off)): 
...  print [(on[i] if choice else off[i]) for i, choice in enumerate(choices)] 
... 
[5.53, 3.12, 3.75, 4.75] 
[5.53, 3.12, 3.75, 1.7] 
[5.53, 3.12, 2.1, 4.75] 
[5.53, 3.12, 2.1, 1.7] 
... <10 more entries omitted ...> 
[1.05, 1.5, 2.1, 4.75] 
[1.05, 1.5, 2.1, 1.7] 

要打印的產品，而不是名單：

>>> import operator 
>>> for choices in itertools.product([0, 1], repeat=len(off)): 
...  elts = [(on[i] if choice else off[i]) for i, choice in enumerate(choices)] 
...  print reduce(operator.mul, elts, 1) 
... 
307.32975 
109.9917 
172.10466 
61.595352 
... 

如果您有可用numpy的，並願意與numpy的陣列，而不是Python列表的工作，然後有一些不錯的工具像numpy.choose可用。例如：

>>> import numpy 
>>> numpy.choose([0, 1, 0, 1], [off, on]) 
array([ 5.53, 1.5 , 3.75, 1.7 ]) 
>>> numpy.product(numpy.choose([0, 1, 0, 1], [off, on])) 
52.880624999999995 

與早期解決方案相結合，得出：

>>> for c in itertools.product([0, 1], repeat=4): 
...  print numpy.product(numpy.choose(c, [off, on])) 
... 
307.32975 
109.9917 
172.10466 
61.595352 
147.7546875 
52.880625 
... 

Answer 2

你們非常接近，itertools.combinations()。

Answer 3

這是你想要的東西：

off = [5.53,3.12,3.75,4.75] 
on = [1.05,1.5,2.1,1.7] 
import itertools as it 
import operator 

indices = list(it.product([0,1],[0,1],[0,1],[0,1])) 
off_on = off,on 
a = [reduce(operator.mul,(off_on[z][i] for i,z in enumerate(x))) for x in indices] 
print a 


#Numpy solution 
import numpy as np 
indices = list(it.product([0,1],[0,1],[0,1],[0,1])) 
off_on = off,on 
b = np.array([off,on]) 
loff = range(len(off)) 
aa = [np.prod(b[list(x),loff]) for x in indices] 

print aa 
print aa == a