i'm a python newbie and i want to check if a each list element is present in another list(while respecting the index) and append this element to a third list. like this. if first element of 'listy'("11-02-jeej") contains first element of list_of_dates ("11-02), i want this element "11-02-jeej" to be appended in the first list of a list of lists. the code below doesn't work for me :(

the output that i want from this code is :[["11-02-jeej"], [2apples], []]
but instead i get : [[], [], []] 

thank you so much !

list_of_dates =["11-02,", "2", "5"]
listy = ["11-02-jeej", "2apples", "d44"]
length = len(list_of_dates)
lst = [[] for m in range(length)]

for i in range(len(list_of_dates)):
    date =  list_of_dates[i]
    for j in range(len(listy)):
         name = listy [j]
    if date in name:
        lst[m].append(name)

print(lst)

CodePudding user response：

There are the following issues in your code:

• The input has a comma in the first string: "11-02,". As you expect this to be a prefix, I suppose that trailing comma should not be there: "11-02"

• The if statement should be inside the inner loop, since it needs the name variable that is assigned there.

• m is not the correct index. It should be i, so you get: lst[i].append(name)

So here is your code with those corrections:

list_of_dates =["11-02", "2", "5"]
listy = ["11-02-jeej", "2apples", "d44"]
length = len(list_of_dates)
lst = [[] for m in range(length)]

for i in range(len(list_of_dates)):
    date =  list_of_dates[i]
    for j in range(len(listy)):
        name = listy [j]
        if date in name:
            lst[i].append(name)

print(lst)

Note that these loops can be written with list comprehension:

lst = [[s for s in listy if prefix in s] for prefix in list_of_dates]

Be aware that for the given example, "2" also occurs in "11-02-jeej", so you have both "11-02" and "2" giving a match, and so that will impact the result. If you wanted "2" to only match with "2apples", then you may want to test a match only at the start of a string, using .startswith().