如何从列表中删除列表

Question

I have created a list where within it I seek to eliminate only the lists whose first value is greater than the second value of it.

I tried creating a second list with the elements to remove but I think it is not the most optimal way.

#y = []

x = [[1, 4], [1, 6], [2, 5], [2, 7], [4, 8], [6, 5], [6, 7], [2, 6], [3, 7], [5, 8], [6, 4], [7, 5]]
for i in range(len(x)):
    if x[i][0] > x[i][1]:
        print(x[i])


#        y.append(x[i])

Is there an optimal way to achieve this?

I want that when printing on screen you get the following:

[[1, 4], [1, 6], [2, 5], [2, 7], [4, 8], [6, 7], [2, 6], [3, 7], [ 5, 8]]

Best regards,

Answer 1

This should work:

y = [[a,b] for a,b in x if a <= b]

Testing:

>>> x = [[1, 4], [1, 6], [2, 5], [2, 7], [4, 8], [6, 5], [6, 7], [2, 6], [3, 7], [5, 8], [6, 4], [7, 5]]
>>> y = [[a,b] for a,b in x if a < b]
>>> y
[[1, 4], [1, 6], [2, 5], [2, 7], [4, 8], [6, 7], [2, 6], [3, 7], [5, 8]]
>>> 

Answer 2

This modifies the original list:

for i, (a, b) in enumerate(x):
    if a > b:
        del x[i]

Answer 3

Creating a new list:

[v for v in x if v[0] <= v[1]]

Or elimination in place:

for i in range(len(x) - 1, -1, -1):  # need to start at the end
    if x[i][0] > x[i][1]:
        x.pop(i)

Answer 4

>>> filtered = list(filter(lambda f: f[0] < f[1], x))
>>> print(filtered)

This will create a new list with the desired values, using the built in filter(function, iterable) function.