Question

我有元组列表（mytuples）和列表列表（mylist）。

我想找出每个列表中mytuples中每个元组出现的次数。

在(2,3)，[1,2,3,4]和[2,3,4,5]中出现元组[2,3]。因此(2,3)的计数为3。

元组和列表大小可以不同。

mytuples = [(2,3), (3,6), (1,2)]
mylist = [[1,2,3,4],[2,3,4,5],[2,3],[4,5,6]]

count={}    
for m in mytuples :
    counter = 0        
    for i in mylist :
        if set(m).issubset(i):
            counter = counter + 1
    count[m]=counter

我的输出是{(2,3):3, (3,6): 0, (1,2):1}

这种方法很好但是当我的列表大小很大时说1000条记录，它更耗时。这可以更快地完成吗？有什么建议吗？

Answer 1

使用 dict理解我们可以将所有内容减少为一行。

假设元组总是成对的：

count = {(x,y):sum((x in i and y in i) for i in mylist) for x,y in mytuples}
# {(1, 2): 1, (2, 3): 3, (3, 6): 0}

如果元组大小未知，您可以使用all()求和：

count = {t:sum(all(x in i for x in t) for i in mylist) for t in mytuples}
# {(1, 2): 1, (2, 3): 3, (3, 6): 0}

如果不清楚：

我们经历了其中的多个：

[all(x in i for x in (2,3)) for i in mylist]
# [True, True, True, False] 
# sum([True, True, True, False]) = 3
# And we assign them back to the tuple

Answer 2

通过小幅调整，您的当前算法可以更快一些：

# Your input data.
tuples = [(2,3), (3,6), (1,2)]
lists = [[1,2,3,4],[2,3,4,5],[2,3],[4,5,6]]

# Convert to sets just once, rather than repeatedly
# within the nested for-loops.
subsets = {t : set(t) for t in tuples}
mainsets = [set(xs) for xs in lists]

# Same as your algorithm, but written differently.
tallies = {
    tup : sum(s.issubset(m) for m in mainsets)
    for tup, s in subsets.items()
}

print(tallies)

Python - 在列表列表中查找元组

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