List comprehension in nested lists Python

Question

dev_a = [[True, 1, 2, 3], [False, 1, 2, 3], [True, 1, 2, 3]]

dev_b = []

My goal is to iterate over nested lists in dev_a and check if [0] is True or False . I need the nested list in dev_b , when [0] is False .

My approach was the following:

if (x[0] for x in dev_a) is False:
   ?

I don't know what to tell the machine, that it will copy the whole dataset.

I need it like this, when it's done:

dev_a = [[True, 1, 2, 3], [False, 1, 2, 3], [True, 1, 2, 3]]

dev_b = [False, 1, 2, 3]

Answer 1

what you want is this:

dev_a = [[True, 1, 2, 3], [False, 1, 2, 3], [True, 1, 2, 3]]

dev_b = [x for x in dev_a if x[0] == False]

take note though, this will be dynamic depending on your input. So you could end up with a single array, or multiple.

Answer 2

for sublist in dev_a:
    if not sublist[0]:
        dev_b.append(sublist)

Something like this?

If you want dev_b to be a flat list, use extend

for sublist in dev_a:
    if not sublist[0]:
        dev_b.extend(sublist)

Answer 3

You could just use filter

dev_a = [[True, 1, 2, 3], [False, 1, 2, 3], [True, 1, 2, 3]]

dev_b = list(filter(lambda l: l[0] is False, dev_a))

print(dev_b)

Result:

[[False, 1, 2, 3]]

Use next() instead of list() if you want just one result instead of a list of lists.

Answer 4

try

dev_a = [[True, 1, 2, 3], [False, 1, 2, 3], [True, 1, 2, 3]]
dev_b = [e for e in dev_a if e[0] is False]
print(dev_b)

output

[[False, 1, 2, 3]]

Answer 5

If you want a flattened list

# filter only the inner items you want using filter function
outer = list(filter(lambda a: a[0] == False, dev_a))
# or without using filter
outer = [a for a in dev_a if not a[0]]

# flatten all items
x = [item for inner in outer for item in inner]

Result:

[False, 1, 2, 3]