根据第二个列表中的元素对列表进行排序和筛选

Question

I have two lists, the first one contains peoples' names, with each person associated with various characters, for example numbers, letters, eg:

listNameAge = ['alain_90xx', 'fred_10y', 'george_50', 'julia_10l','alain_10_aa', 'fred_90', 'julia_50', 'george_10s', 'alain_50', 'fred_50', 'julia_90']

The second one contains the name of the person:

listName = ['fred', 'julia', 'alain', 'george']

Using the second list, I would like to associate a third list to the first one, such that each name in the first list is associated with its index position in the second one, ie:

thirdlist = [2, 0, 3, 1, 2, 0, 1, 3, 2, 0, 1]

The name and characters are separated by an underscore, but the character can be of any sort. I could loop over the elements of listNameAge , separate the name of the persons from the rest of the characters using a .split('_') on the string, find which name it is and find its index in listName using a second loop.

I was however wondering if there is a simpler way to do this, ie avoid using loop and use only a comprehension list?

Answer 1

While you can do this with a one-liner, I think that, for efficiency, it wold pay to build a dictionary:

namePos = dict((name, i) for (i, name) in enumerate(listName))
>>> [namePos[n.split('_')[0]] for n in listNameAge]
[2, 0, 3, 1, 2, 0, 1, 3, 2, 0, 1]

The (expected) running time of this code, is Θ(m + n) where m is the length of the first list, and n the length of the other one.

Answer 2

For this question in specific, I would recommend you use a loop just for clarity. However, if you must use a list comprehension, you can do that essentially the same way:

thirdlist = [listName.index(x[:x.find('_')]) for x in listNameAge]

Answer 3

thirdList = [listName.index(string.split("_")[0]) for string in listNameAge]

它是由listName.index(string.split("_")[0]组成的列表理解，其中为listNameAge每个项定义string string.split("_")[0]是从开头的字符串第一个下划线的字符串，所以listName.index(string.split("_")[0]是listName第一次出现的字符串。

Answer 4

You can take each item in listNameAge , then split on '_' , get the first part of the split, then use index to find it in the second list.

>>> [listName.index(i.split('_')[0]) for i in listNameAge]
[2, 0, 3, 1, 2, 0, 1, 3, 2, 0, 1]

Answer 5

You can try this , check whenever the listNameAge appears in listName :

for x in listNameAge:
    for y in listName:
        if y in x:
            thirdList.append(listName.index(y))

result :

[2, 0, 3, 1, 2, 0, 1, 3, 2, 0, 1]

Answer 6

I would strongly recommend against using .index() as its complexity is O(n) and makes the overall complexity of this operation O(mn) where m and n are sizes of the lists.

Here's a fast one liner using generators:

map(lambda (x,y): y[x[:x.find('_')]],izip(listNameAge, repeat(dict(izip(listName, count())))))

More readable version would be (as Ami has shown):

nameMap = dict(izip(listName, xrange(len(listName))))
thirdList = map(lambda x: nameMap[x[:x.find('_')]],listNameAge)