How to convert this to a list comprehension

Question

I want to convert this part of the code to a list comprehension, however my knowledge of this method is still very weak and it does not occur to me how, if you could help me, I would appreciate it.

list_1 = ["h",'i',' ', 'w','o','r','l','d'] 
list_2 = ["h",'i',' ', 'm','o','o','n']   

list_3 = []

for word in list_1:
    if word in list_2 and word not in list_3:
        list_3.append(word)

print(list_3)

Your question had some translation errors so I fixed them. Hope that helps! If I misunderstood please let me know. — wjandrea, Mar 22 '20 at 20:18
@Errol I'm inferring that it already works properly, but OP just wants to convert it to a comprehension. — wjandrea, Mar 22 '20 at 20:26

Rado Cisar · Answer 1 · 2020-03-24T10:14:32.197

2

Here you go:

[list_3.append(w) for w in list_1 if (w in list_2) and (w not in list_3)]
list_3

Output:

['h', 'i', ' ', 'o']

Enjoy!

edited Mar 24 '20 at 10:14

answered Mar 22 '20 at 20:39

Rado Cisar

21
4

This works, but the style is bad. [Don't use comprehensions for side effects](https://stackoverflow.com/q/5753597/4518341), and [don't use bitwise operators when you should use logical operators](https://stackoverflow.com/q/3845018/4518341). – wjandrea Mar 22 '20 at 20:43
@wjandrea yes, my method may be a bit computationally expensive, but it solves the example given. Unless you have a complex code for a large program, this should not be a problem. for the bitwise vs boolean operators, you are right the boolean `and` makes more sense here. – Rado Cisar Mar 24 '20 at 10:22

wjandrea · Accepted Answer · 2020-03-22T20:35:18.627

1

In this case, this works:

list_3 = [word for word in list_1 if word in list_2]
print(list_3)  # -> ['h', 'i', ' ', 'o']

If you want to keep excluding duplicates in list_3 then it gets a bit more complicated. Check out Removing duplicates in lists. For example, this will work in Python 3.7+:

dict_3 = {word: None for word in list_1 if word in list_2}
print(list(dict_3))  # -> ['h', 'i', ' ', 'o']

edited Mar 22 '20 at 20:35

answered Mar 22 '20 at 20:21

wjandrea

28,235
9
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thanks bro do not notice that I have to use the if condition in these cases – epa la arepa Mar 22 '20 at 20:30
@wjandrea This is not the correct answer, it breaks for an input like `list_1 = list("hi there")` and `list_3` will include duplicates... – Guimoute Mar 22 '20 at 20:34
@Guimoute Added a solution for that – wjandrea Mar 22 '20 at 20:35
bro but here elements are missing – epa la arepa Mar 22 '20 at 20:47
@epalaarepa What do you mean? – wjandrea Mar 22 '20 at 20:50
['h', 'i', ' ', 'o'] ['h', 'i', ' ', 'w', 'm', 'r', 'l', 'd', 'o'] – epa la arepa Mar 22 '20 at 20:55
the second list is the one that should quit the function – epa la arepa Mar 22 '20 at 20:55
@epalaarepa What function? What do you mean by "quit"? – wjandrea Mar 22 '20 at 20:57
['h', 'i', ' ', 'w', 'm', 'r', 'l', 'd', 'o'] the purpose of the function is to generate a new list and put non-repeating elements inside the new list – epa la arepa Mar 22 '20 at 21:04
@epalaarepa The code in your question doesn't do that. I'm not a mind-reader... Ask a new question if you need. – wjandrea Mar 22 '20 at 21:14
but in the code it is seen in the `code if word in list_2 and word not in list_3: ` check and add the words that are not repeated – epa la arepa Mar 22 '20 at 21:40

score 1 · Answer 3 · answered Mar 22 '20 at 20:42

1

The right way to do it is using set data structure.

list_1 = ["h",'i',' ', 'w','o','r','l','d'] 
list_2 = ["h",'i',' ', 'm','o','o','n'] 
set_1 = set(list_1)
set_2 = set(list_2)
set_3 = set_1 & set_2
print(set_3)

output

{' ', 'o', 'i', 'h'}

answered Mar 22 '20 at 20:42

balderman

22,927
7
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Alex Liu 33214 · Answer 4 · 2020-03-22T20:15:05.133

-1

This is all you need

next time please do more reaserach

list_1 = ["h",'i',' ', 'w','o','r','l','d'] 
list_2 = ["h",'i',' ', 'm','o','o','n','']   

list_3 = []

for i in range(0, len(list_1)):
    if list_1[i] != list_2[i]
        list_3.append(list_1[i])

print(list_3)

result:

['m','o','n']

edited Mar 22 '20 at 20:15

answered Mar 22 '20 at 20:02

Alex Liu 33214

72
8

2

that's list comprehension? – epa la arepa Mar 22 '20 at 20:07
@epalaarepa It's what your code wanted to do, I can explain – Alex Liu 33214 Mar 22 '20 at 20:09
@epalaarepa this code scans through both lists one by one and every time the letter is different, a boolean for the if statement returns true and appends to the new list. – Alex Liu 33214 Mar 22 '20 at 20:11
1

I thought of something like this list_3 = [ word for word in list_1 and word not in list_2 ] – epa la arepa Mar 22 '20 at 20:12
@epalaarepa The logic is coorect but the syntax had problems – Alex Liu 33214 Mar 22 '20 at 20:15
NameError: name 'word' is not defined gives me this error – epa la arepa Mar 22 '20 at 20:16
@epalaarepa when you use your code, word is just like my i, a variable that goes through 0 to 8 because your string is 8 objects long – Alex Liu 33214 Mar 22 '20 at 20:18
If not, why if the letter according to what I understand is the one that is saved in the new list, and the program says that the letter is not defined – epa la arepa Mar 22 '20 at 20:18
@epalaarepa that is because of the fact that list_1 is 8 long and list_2 is 7 long so when word goes to 8, the whole thing crashes – Alex Liu 33214 Mar 22 '20 at 20:21
1

your answer is fine but what I need is to reduce the size of the code for that is the understanding of lists as well as I put the code it runs fine – epa la arepa Mar 22 '20 at 20:22
@epalaarepa wjandrea's answer is more strait forward, mabe go check out his? – Alex Liu 33214 Mar 22 '20 at 20:22
@AlexLiu10101 Sorry, I should have mentioned, OP's looking for a list comprehension. The question had a translation problem that I fixed, "understanding" -> "comprehension". – wjandrea Mar 22 '20 at 20:29

How to convert this to a list comprehension

4 Answers4