从嵌套列表中删除子列表（子列表的查找索引）

Question

I have a nested list like this:

lists = [[['L', 5], ['B', 20], ['A', 10]], 
        [['B', 200], ['J', 90]], 
        [['L', 5], ['L', 6]], 
        [['A', 10], ['L', 12], ['A', 11], ['A', 15]]]

How can I remove sublists that have string which is not A,B,L(remove the whole sublist not only the list that not A,B,L) How can I find the index of sublist that invalid items belong to (in this case is 1)(need the index for further task)

This is what I have tried, it can find invalid item but I do not know to find index of the sublist

for j in range (len(lists)):
    for i in range (len(lists[j])):
        if lists[j][i][0] != 'L' and lists[j][i][0] != 'A' and lists[j][i][0] != 'B':
            return False
return True 

I want the result to be like this:

lists = [[['L', 5], ['B', 20], ['A', 10]],  
        [['L', 5], ['L', 6]], 
        [['A', 10], ['L', 12], ['A', 11], ['A', 15]]]

Answer 1

You can efficiently modify lists inplace using the reverse-delete idiom:

keep = ('A', 'B', 'L')
for i in reversed(range(len(lists))):
    if any(l[0] not in keep for l in lists[i]):  
        del lists[i]

print(lists)
# [[['L', 5], ['B', 20], ['A', 10]],
#  [['L', 5], ['L', 6]],
#  [['A', 10], ['L', 12], ['A', 11], ['A', 15]]]

any returns True if any of the sublists' first element is not in keep .

---

Alternatively, you can create a new list with a list comprehension:

[l for l in lists if not any(l_[0] not in keep for l_ in l)]
# [[['L', 5], ['B', 20], ['A', 10]],
#  [['L', 5], ['L', 6]],
#  [['A', 10], ['L', 12], ['A', 11], ['A', 15]]]

Answer 2

As @coldspeed suggests, using a set to check if the letters exists allows optimal O(1) lookups.

If you don't want to use any builtin functions like any() , first make a function that checks if the first letter of each sublists inner list exists:

valid = {"A", "B", "L"}

def check_valid(sublst):
    for fst, *_ in sublst: 
        if fst not in valid:
            return False
    return True

Or without tuple unpacking if you prefer:

def check_valid(sublst):
    for lst in sublst:
        if lst[0] not in valid:
            return False
    return True

Then you can reconstruct a new list with the incorrect lists filtered out:

result = []
for sublst in lists:
    if check_valid(sublst):
        result.append(sublst)

print(result)
# [[['L', 5], ['B', 20], ['A', 10]], [['L', 5], ['L', 6]], [['A', 10], ['L', 12], ['A', 11], ['A', 15]]]

Or as a list comprehension:

result = [sublst for sublst in lists if check_valid(sublst)]
print(result)
# [[['L', 5], ['B', 20], ['A', 10]], [['L', 5], ['L', 6]], [['A', 10], ['L', 12], ['A', 11], ['A', 15]]]

Note: It is always better to use builtin functions for convenience, since it saves you having to reinvent the wheel and usually leads to shorter, more concise code.