Enumerate a list based in a condition in other list-CodePudding

list1 is the original list
list2 is the criterion list
list3 is the resulting list

Operation: if index (first element of each nested list) in list1 occurs in list2, delete this nested list and continue enumeration with the next element not in list2.

Example 1:

list1 = [[1,'a'],[2,'b'],[3,'c'],[4,'d'],[5,'e'],[6,'f'],[7,'g'], [8,'h']]
list2 = [3,4,5,7]
list3 = [[1,'a'],[2,'b'],[3,'f'],[4,'h']]

Example 2:

list1 = [[0,'do'],[1,'re'],[2,'mi'],[3,'fa'],[4,'sol'],[5,'la'],[6,'si']]
list2 = [1,3,5]
list3 = [[0,'do'],[1,'mi'],[2,'sol'],[3,'si']]

CodePudding user response：

Build two different iterators -- one with the unfiltered first elements, one with the second elements filtered according to the first element and list2. Zip them together to build list3.

>>> list1 = [[1,'a'],[2,'b'],[3,'c'],[4,'d'],[5,'e'],[6,'f'],[7,'g'], [8,'h']]
>>> list2 = [3,4,5,7]
>>> [list(z) for z in zip(next(zip(*list1)), (j for i, j in list1 if i not in list2))]
[[1, 'a'], [2, 'b'], [3, 'f'], [4, 'h']]

CodePudding user response：

Try:

list1 = [
    [0, "do"],
    [1, "re"],
    [2, "mi"],
    [3, "fa"],
    [4, "sol"],
    [5, "la"],
    [6, "si"],
]
list2 = [1, 3, 5]

out = [b for a, b in list1 if a not in list2]
out = [[a, c] for (a, _), c in zip(list1, out)]
print(out)

Prints:

[[0, 'do'], [1, 'mi'], [2, 'sol'], [3, 'si']]

CodePudding user response：

You can use list comprehensions like so:

list1 = [[1,'a'],[2,'b'],[3,'c'],[4,'d'],[5,'e'],[6,'f'],[7,'g'], [8,'h']]
list2 = [3,4,5,7]

temp = [i for i in list1 if not i[0] in list2]
list3 = [[i 1, temp[i][1]] for i in range(len(temp))]
print(list3)

Output:

[[1, 'a'], [2, 'b'], [3, 'f'], [4, 'h']]

CodePudding user response：

Obviously there are plenty of ways to make this happen, here's one.

def yourFunctionName(list1,list2):
    result = []
    n = 1
    for element in list1:
        if not element[0] in list2:
            result.append([n,element[1]])
    return result

CodePudding user response：

You can accomplish this with a list comprehension, using list1 and list2 from the first example:

list3 = [[i list1[0][0], elem[1]] for i, elem in enumerate(e for e in list1 if e[0] not in list2)]
print(list3)

The idea is that I get rid of all the lists in the criterion list first, and then I re-enumerate them based on the first element in list1.

Output:

[[1, 'a'], [2, 'b'], [3, 'f'], [4, 'h']]