如何检查列表中的所有元素是否都包含在其他元素中

Question

I've two list and need to check if all elements of list_1 are contained in list_2 id not, save those elements in a new list.

SO I've this:

list_1 = ['item','item','item']
list_2 = ['item_2','item_2','item_2']
list_3 = []

for i in range(len(list_1)):
    flag = True
    aux = list_1[i]
    for j in range(len(list_2)):
        if aux == list_2[j]:
            flag == False
            break
    if flag:
        list_3.append(aux)

But this is very slow, there is a way to improve the speed?

maybe some built pandas function?

Edit.

I don't need pandas to this, but the list are infact columns two Data Frames, I just write it with list because it's a more general case.

Answer 1

IIUC I think you could use isin and all methods of pd.Series :

import pandas as pd
l1, l2 = map(pd.Series, [list_1, list_2])

In [3]: l1      
Out[3]:         
0    item       
1    item       
2    item       
dtype: object   

In [4]: l2      
Out[4]:         
0    item_2     
1    item_2     
2    item_2     
dtype: object   

In [5]: l1.isin(l2)
Out[5]:
0    False
1    False
2    False
dtype: bool

In [6]: l1.isin(l2).all()
Out[6]: False

Answer 2

You seem to be interested in the union and difference of sets .

You could check for these, like this:

a = set([1, 5, 6, 0])
b = set([0, 8, 2, 3, -5])
a.difference(b)   # returns: {1, 5, 6}
a.intersection(b) # returns {0}
b.difference(a)   # returns: {-5, 2, 3, 8}

Answer 3

如果要在list_1中找到不在list_2的元素并将其保存在其他列表中（在本例中为list_3 ，则可以使用列表 list_3

>>> list_3 = [i for i in list_1 if i not in list_2]

Answer 4

I find numba to be fun for loops, it really brings up the speed with minimum effort:

from numba import jit
@jit
def jitstack():
    list_1 = ['item','item','item']
    list_2 = ['item_2','item_2','item_2']
    list_3 = []
    for i in range(len(list_1)):
        flag = True
        aux = list_1[i]
        for j in range(len(list_2)):
            if aux == list_2[j]:
                flag == False
                break
        if flag:
            list_3.append(aux)

Timed with timeit in iPython notebook: https://drive.google.com/file/d/0B0KNIF4xMP3UNW93LWlmWUFqbnc/view?usp=sharing

Original: 3.72 µs per loop

Numba: 22.4 ns per loop

and also the list_3 = [i for i in list_1 if i not in list_2]: 1.22 µs per loop