如何将 pop 用于列表列表

Question

def process_timecards():
    timecards = []
    with open("timecards.txt") as f:
        reader = csv.reader(f)
        listoftimecards = [list(map(float,row)) for row in reader]
    print(listoftimecards)
    list1 = listoftimecards.pop(0)
    print(list1)

[[688997.0, 5.0, 6.8, 8.0, 7.7, 6.6, 5.2, 7.1, 4.0, 7.5, 7.6], [939825.0, 7.9, 6.6, 6.8, 7.4, 6.4, 5.1, 6.7, 7.3, 6.8, 4.1], [900100.0, 5.1, 6.8, 5.0, 6.6, 7.7, 5.1, 7.5], [969829.0, 6.4, 6.6, 4.4, 5.0, 7.1, 7.1, 4.1, 6.5], [283809.0, 7.2, 5.8, 7.6, 5.3, 6.4, 4.6, 6.4, 5.0, 7.5], [224568.0, 5.2, 6.9, 4.2, 6.4, 5.3, 6.8, 4.4], [163695.0, 4.8, 7.2, 7.2, 4.7, 5.1, 7.3, 7.5, 4.5, 4.6, 7.0], [454912.0, 5.5, 5.3, 4.5, 4.3, 5.5], [285767.0, 7.5, 6.5, 6.3, 4.7, 6.8, 7.1, 6.6, 6.6], [674261.0, 7.2, 6.2, 4.9, 6.5, 7.2, 7.5, 5.0, 7.9], [426824.0, 7.4, 6.5, 5.7, 8.0, 6.9, 7.5, 6.5, 7.5], [934003.0, 5.8, 7.5, 5.8, 4.8, 5.9, 4.8, 4.0, 6.6, 5.5, 7.2]]

This is the list of lists that I have, and i need to grab the first value of each list inside of the list of lists and store that into a list.

I thought i could use pop, but that only goes to the first list. It results in just printing out the first value of the list, which is the first list.

Any advice? I was thinking maybe a for loop, but i have no idea how i would format it.

Answer 1

list1 = [sublist[0] for sublist in listoftimecards]

Answer 2

since you want only the first element you could grab from each row only the first element using the built-in function next :

def process_timecards():
    timecards = []
    with open("timecards.txt") as f:
        reader = csv.reader(f)
        list1 = [next(map(float,row)) for row in reader]
    print(list1)

Answer 3

list1 = []
for item in listoftimecards:
    list1.append(item[0]) 

This loops through each item in listoftimecards and appends the first item of every list in listoftimecards to list1

This code below does the same thing as the code above.

list1 = [x[0] for x in listoftimecards]

Answer 4

In Python 2, you can use:

map(lambda lst:lst[0], listoftimecards)

In Python 3, map returns you a generator so you need to call list():

list(map(lambda lst:lst[0], listoftimecards))

but if you only wish to iterate the result using map will be memory efficient.