如何在python中基于另一个列表重新排列列表

alvin Christianto • 5 年前 • 1514 次点击

我想根据以下条件重新排列列表(或创建新列表):

fixedList = ['116','117','114','99','102','101','95']

unorderedList = ['99','116','117']

我想列个新单子,比如:

newList = ['116','117','99'] .

新列表 基于 ,并且排序基于 固定列表

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文章 [ 5 ] | 最新文章 5 年前

• 1 楼

Nithin R 5 年前

import numpy as np

fixedList = ['116','117','114','99','102','101','95']
unorderedList = ['99','116','117']

# isin returns a boolean array of the same shape as element 
# that is True where an element of unorderedList is in fixedList and False otherwise.

mask = np.isin(np.array(fixedList), np.array(unorderedList))

#Now form new list using mask
newlist = (np.array(fixedList)[mask]).tolist()

现在新列表包含以下内容

参考 https://docs.scipy.org/doc/numpy/reference/generated/numpy.isin.html

• 2 楼

Sayed Hussain Mehdi 5 年前

def intersection(lst1, lst2): 
    lst3 = [value for value in lst1 if value in lst2] 
    return lst3

fixedList = ['116','117','114','99','102','101','95']
unorderedList = ['99','116','117']

print( intersection(fixedList,unorderedList))

https://www.geeksforgeeks.org/python-intersection-two-lists/

• 3 楼

fmarm 5 年前

[el for el in fixedList if el in unorderedList]

结果如期而至

['116', '117', '99']

• 4 楼

Alexander Lekontsev 5 年前

像这样的东西。

sorted(unorderedList, key=fixedList.index)

• 5 楼

jarmod 5 年前

使用自定义排序,其中要排序的键是固定列表中的索引,例如:

fixedList = ['116','117','114','99','102','101','95']
unorderedList = ['99','116','117']
newList = sorted(unorderedList, key=lambda x:fixedList.index(x))

输出为:

['116', '117', '99']

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