列表中列表上的列表理解

Question

If I create two lists that contain lists like this:

bad_list.append(['blue_widget', 'cracked', '776'])
bad_list.append(['red_widget', 'not_smooth', '545']) 
bad_list.append(['yellow_widget', 'spots', '35']) 
bad_list.append(['green_widget', 'smells_bad', '10'])
bad_list.append(['purple_widget', 'not_really_purple', '10'])


good_list.append(['blue_widget', 'ok', '776'])
good_list.append(['red_widget', 'ok', '545']) 
good_list.append(['green_widget', 'ok', '10'])

I would love to be able to use list comprehension to compare the two lists and remove all items in the bad list that are present in the good list using the first element (x_widget) as item to compare. Using the example above I should be left with:

['yellow_widget', 'spots', '35']
['purple_widget', 'not_really_purple', '10']

I have tried using list comprehension and it works but the new list does not retain each line:

final_list = [x for x in bad_list[0] if x not in good_list[0]]

When I print out the contents using for item in final_list I get something like:

yellow_widget
smells_bad
10

Any clues would be much appreciated.

Answer 1

One liner

[x for x in bad_list if any(x[0] == y[0] for y in good_list)]

*thanks @Bakuriu

Answer 2

Not really optimized by any means, but this should work: http://codecube.io/AD7RHA

bad_list=[]
good_list=[]

bad_list.append(['blue_widget', 'cracked', '776'])
bad_list.append(['red_widget', 'not_smooth', '545']) 
bad_list.append(['yellow_widget', 'spots', '35']) 
bad_list.append(['green_widget', 'smells_bad', '10'])
bad_list.append(['purple_widget', 'not_really_purple', '10'])


good_list.append(['blue_widget', 'ok', '776'])
good_list.append(['red_widget', 'ok', '545']) 
good_list.append(['green_widget', 'ok', '10'])

# ['yellow_widget', 'spots', '35']
# ['purple_widget', 'not_really_purple', '10']

labels = zip(*good_list)[0]

new_bad_list=[]

for item in bad_list:
    if item[0] not in labels:
        new_bad_list.append(item)

print new_bad_list

or this one-liner:

new_bad_list=[item for item in bad_list if item[0] not in zip(*good_list)[0]]

Answer 3

Try This:

print [ele for ele in bad_list if ele[0] not in [i[0] for i in good_list]]

Output:

[['yellow_widget', 'spots', '35'], ['purple_widget', 'not_really_purple', '10']]

Answer 4

There is a more efficient solution. Make a set from your list

bad_set = set(bad_list)
good_set = set(good_list)

Now to remove all items in the bad list that are present in the good list, you can simple substract the sets:

bad_set - good_set

Convert set back to list if you like.

Answer 5

the simplest way is:

final_list = [x for x in bad_list if x[0] not in [x[0] for x in good_list]]

but notice that to test an element's existing in a list is not that efficient.

So, you can first build a set:

good_list_names = set([x[0] for x in good_list])

and then:

final_list = [x for x in bad_list if x[0] not in good_list_names]