问题描述
这是我设法在R中工作的东西。出于某种原因，现在它必须在python中。 翻译不是直接的。
如果我有两个列表，一个是分子，一个是分母，那么如何挑选两个比率的所有集合，以使每个集合中两个比率的分子不相同。 同样，我不希望分母相同。
upList = ("up1","up2","up3")
downList = ("down1","down2","down3")
我想生产：
up1down1, up2down2
up1down1, up2down3
up1down1, up3down2
up1down1, up3down3
up1down2, up2down1
up1down2, up2down3
up1down2, up3down1
up1down2, up3down3
等等 ...
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])))
这给了我4个分子/分母对的集合，我不知道如何以分子不相同的方式组合这4对。 上面的代码产生许多行，例如：
[('up1', 'down1'), ('up1', 'down2'), ('up2', 'down1'), ('up2', 'down2')]
在这行之外，我想生产
[('up1', 'down1'), ('up2', 'down2')]
[('up1', 'down2'), ('up2', 'down1')]
1楼
编辑：纠正，感谢@Blckknght的评论。
给定您的输入数据：
upList = ("up1","up2","up3")
downList = ("down1","down2","down3")
您要首先创建分子x分母的所有排列。 可从 。
from itertools import product
ratios = product(upList, downList)
接下来，您要查找产品中2个不同项目的所有组合。 这是2组合，可以通过获得。
from itertools import combinations
ratio_pairs = combinations(ratios, 2)
但是，您想将组合限制为两个项不共享相同分子，也不共享相同分母的组合。 这是一个过滤列表的理解：
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'))
2楼
我觉得你很亲密。
您的代码可以正确工作，直到您创建要打印的最终列表。
我想您想为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)])
这不会完全符合您想要的顺序，但是它会产生所有所需的对组合。
如果您需要按地志顺序排列结果，请将结果放在列表中并进行sort 。
假定您不关心列表中的对的顺序。
如果这样做，您将需要另外两个结果，并且上面的对将交换为[(b, d), (a, c)]和[(b, c), (a, d)] ）。
3楼
您可以像这样通过itertools使用product方法，以得到所需的输出：
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
...