在嵌套列表中查找索引

发布于 2024-12-11 12:22:51 字数 469 浏览 0 评论 0原文

可能的重复：
Python 中的高效单词索引

对基本问题表示歉意，如何找到索引< code>((1,2),(2,1),(2,2)) 对应于 ref 中的“yellow”并存储它们，以便我可以在事后访问它们而无需重新运行搜索？

ref1 = "this is a test" 
ref2 = "this is a yellow test"
ref3 = "this is yellow"
ref4 = "yellow and orange are colors"
ref = ((ref1,ref2),(ref3,ref4))

原文

Possible Duplicate:
efficient word indexing in python

Apologies for the basic question, how can I find the indexes ((1,2),(2,1),(2,2)) corresponding to "yellow" in ref and store them so that I can access them afterwords without rerunning the search?

ref1 = "this is a test" 
ref2 = "this is a yellow test"
ref3 = "this is yellow"
ref4 = "yellow and orange are colors"
ref = ((ref1,ref2),(ref3,ref4))

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过度放纵 2024-12-18 12:22:51

只是为了它：

[(k,i) for k,m in enumerate(ref, 1) for i,j in enumerate(m, 1) if 'yellow' in j]

但您可能想要真正的索引（从零开始）...如果是这样，请删除 enumerate 的第二个参数。

Just for the sake of it:

[(k,i) for k,m in enumerate(ref, 1) for i,j in enumerate(m, 1) if 'yellow' in j]

but you probably want the real index (starting at zero)... Remove the second argument of enumerate if so.

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空名 2024-12-18 12:22:51

我不确定你的意思，但是如果你想在每次看到“黄色”这个词时从嵌套列表（或元组）的列表（或元组）中生成一个对的列表，那么：

answer = []
for i in range(len(ref)):
    nested_tuple = ref[i]
    for j in range(len(nested_tuple)):
        string = ref[i][j]
        if string.find('yellow') != -1: answer.append((i,j))

I'm not sure what you mean, but if you want to generate a list of pairs from a list (or tuple) of nested lists (or tuples) for every time you see the word 'yellow' then:

answer = []
for i in range(len(ref)):
    nested_tuple = ref[i]
    for j in range(len(nested_tuple)):
        string = ref[i][j]
        if string.find('yellow') != -1: answer.append((i,j))

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