Hi All,
I have the below code:
list_1 = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]
win_1 = ["1", "2", "3", "4"]
if win_1 in list_1:
print("yes")It doesn't print anything, how do I check if a list is in another list?
Not sure what you want but, this will print anything from win_1 that is in list_1
list_1 = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]
win_1 = ["1", "2", "3", "4"]
for item in list_1:
if item in win_1:
print(item)Output:
1
2
3
4
list_1 = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]
win_1 = ["1", "2", "3", "4"]
win_2 = ['1', '12']
print(all(item in list_1 for item in win_1))
print(all(item in list_1 for item in win_2))
Output:
True
False
Membership testing in a set is faster than membership testing in a list. However, converting list to set also costs some time. So it's up to specific conditions and terms what to prefer.
list_1 = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]
win_1 = ["1", "2", "3", "4"]
print(set(win_1).issubset(list_1)) # are all win_1 elements in list_1
print(set(win_1).intersection(list_1)) # what elements of win_1 in list_1
Output:
True
{'4', '3', '2', '1'}
Looks like you're asking about a subsequence. You can check if a slice of a list matches your subsequence.
list_1 = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]
win_1 = ["1", "2", "3", "4"]
if any(win_1 == list_1[x:x+len(win_1)] for x in range(len(list_1) - len(win_1) + 1)):
print("yes")List slicing and list comparison can take a while. So this can be expensive as the lists get long.