Question

这是我设法在R中工作的东西。现在它需要在python中，原因。翻译不是直截了当的。

如果我有两个列表，一个分子和一个分母，那么如何选择所有两个比率的集合，使得每个集合中两个比率的分子不相同。同样，我也不希望分母是一样的。

upList = ("up1","up2","up3")
downList = ("down1","down2","down3")

我想制作：

up1down1, up2down2
up1down1, up2down3
up1down1, up3down2
up1down1, up3down3

up1down2, up2down1
up1down2, up2down3
up1down2, up3down1
up1down2, up3down3

依旧......

我最初的尝试涉及itertools.product

upSets = itertools.combinations(upList,2)
downSets = itertools.combinations(downList,2)

allSets = itertools.product(upSets,downSets)

for aset in allSets:
    print (aset)
    print (list(itertools.product(geneSet[0],geneSet[1])))

这让我获得了4个分子/分母对的集合，我无法弄清楚如何将这4对结合起来，使得分子不是相同的。上面的代码产生了许多行，如：

[('up1', 'down1'), ('up1', 'down2'), ('up2', 'down1'), ('up2', 'down2')]

出于这一行，我想制作

[('up1', 'down1'), ('up2', 'down2')]
[('up1', 'down2'), ('up2', 'down1')]

Answer 1

修改：已更正，感谢@Blckknght的评论。

根据您的输入数据：

upList = ("up1","up2","up3")
downList = ("down1","down2","down3")

您想首先创建分子x分母的所有排列。这可以从itertools.product获得。

from itertools import product

ratios = product(upList, downList)

接下来，您需要查找产品中2个不同商品的所有组合。这是一个2组合，可以通过itertools.combinations获得。

from itertools import combinations

ratio_pairs = combinations(ratios, 2)

但是您希望将组合限制为两个项目不共享相同分子的组合，也不要将它们共享相同的分母。这是一个过滤列表理解：

distinct_ratio_pairs = [ (p1,p2) for p1,p2 in ratio_pairs if p1[0] != p2[0] and p1[1] != p2[1] ]

for drp in distinct_ratio_pairs:
    print(drp)

输出：

(('up1', 'down1'), ('up2', 'down2'))
(('up1', 'down1'), ('up2', 'down3'))
(('up1', 'down1'), ('up3', 'down2'))
(('up1', 'down1'), ('up3', 'down3'))
(('up1', 'down2'), ('up2', 'down1'))
(('up1', 'down2'), ('up2', 'down3'))
(('up1', 'down2'), ('up3', 'down1'))
(('up1', 'down2'), ('up3', 'down3'))
(('up1', 'down3'), ('up2', 'down1'))
(('up1', 'down3'), ('up2', 'down2'))
(('up1', 'down3'), ('up3', 'down1'))
(('up1', 'down3'), ('up3', 'down2'))
(('up2', 'down1'), ('up3', 'down2'))
(('up2', 'down1'), ('up3', 'down3'))
(('up2', 'down2'), ('up3', 'down1'))
(('up2', 'down2'), ('up3', 'down3'))
(('up2', 'down3'), ('up3', 'down1'))
(('up2', 'down3'), ('up3', 'down2'))

Answer 2

我认为你非常接近。在您创建要打印的最终列表之前，您的代码工作正常。我想你想为allSets产生的每个项目创建两个列表：

upSets = itertools.combinations(upList,2)
downSets = itertools.combinations(downList,2)

allSets = itertools.product(upSets,downSets)

for (a, b), (c, d) in allSets:
    print([(a, c), (b, d)])
    print([(a, d), (b, c)])

这不符合您想要的顺序，但它应该产生所有想要的对组合。如果您需要按字母顺序排列结果，请将结果放在一个列表中sort。

假设您不关心列表中对的顺序。如果您这样做，那么您需要在上面交换的对（[(b, d), (a, c)]和[(b, c), (a, d)]）中再获得两个结果。

Answer 3

您可以像product这样使用itertools方法，以获得所需的输出：

from itertools import product

a = ("up1","up2","up3")
# assuming your b variable is like this one
b = ("down1", "down2","down3")

c = ["".join(k) for k in list(product(a,b))]

subfinal = list(product(c,c))
# removing the duplicates
# maybe not the best way to do it...
# also removing those kind of data: up1down1,up1down2
# also removing those kind of data: up1down1,up3down1
final = [k for k in subfinal if k[0] != k[1] and k[0][:3] != k[1][:3] and k[0][3:] != k[1][3:]]
print('total: ', len(final))
for k in final:
    print(", ".join(k))

输出：

total:  36
up1down1, up2down2
up1down1, up2down3
up1down1, up3down2
up1down1, up3down3
up1down2, up2down1
up1down2, up2down3
up1down2, up3down1
up1down2, up3down3
...

列表分子和分母的组合

3 个答案: