如何删除列表中的最后一项？

I have this program that calculates the time taken to answer a specific question, and quits out of the while loop when answer is incorrect, but i want to delete the last calculation, so i can call min() and it not be the wrong time, sorry if this is confusing.

from time import time

q = input('What do you want to type? ')
a = ' '
record = []
while a != '':
    start = time()
    a = input('Type: ')
    end = time()
    v = end-start
    record.append(v)
    if a == q:
        print('Time taken to type name: {:.2f}'.format(v))
    else:
        break
for i in record:
    print('{:.2f} seconds.'.format(i))

If I understood the question correctly, you can use the slicing notation to keep everything except the last item:

record = record[:-1]

But a better way is to delete the item directly:

del record[-1]

Note 1: Note that using record = record[:-1] does not really remove the last element, but assign the sublist to record. This makes a difference if you run it inside a function and record is a parameter. With record = record[:-1] the original list (outside the function) is unchanged, with del record[-1] or record.pop() the list is changed. (as stated by @pltrdy in the comments)

Note 2: The code could use some Python idioms. I highly recommend reading this:
Code Like a Pythonista: Idiomatic Python (via wayback machine archive).

you should use this

del record[-1]

The problem with

record = record[:-1]

Is that it makes a copy of the list every time you remove an item, so isn’t very efficient

list.pop() removes and returns the last element of the list.

You need:

record = record[:-1]

before the for loop.

This will set record to the current record list but without the last item. You may, depending on your needs, want to ensure the list isn’t empty before doing this.

If you do a lot with timing, I can recommend this little (20 line) context manager:

• https://github.com/brouberol/timer-context-manager

You code could look like this then:

#!/usr/bin/env python
# coding: utf-8

from timer import Timer

if __name__ == '__main__':
    a, record = None, []
    while not a == '':
        with Timer() as t: # everything in the block will be timed
            a = input('Type: ')
        record.append(t.elapsed_s)
    # drop the last item (makes a copy of the list):
    record = record[:-1] 
    # or just delete it:
    # del record[-1]

Just for reference, here’s the content of the Timer context manager in full:

from timeit import default_timer

class Timer(object):
    """ A timer as a context manager. """

    def __init__(self):
        self.timer = default_timer
        # measures wall clock time, not CPU time!
        # On Unix systems, it corresponds to time.time
        # On Windows systems, it corresponds to time.clock

    def __enter__(self):
        self.start = self.timer() # measure start time
        return self

    def __exit__(self, exc_type, exc_value, exc_traceback):
        self.end = self.timer() # measure end time
        self.elapsed_s = self.end - self.start # elapsed time, in seconds
        self.elapsed_ms = self.elapsed_s * 1000  # elapsed time, in milliseconds

just simply use list.pop() now if you want it the other way use : list.popleft()

If you have a list of lists (tracked_output_sheet in my case), where you want to delete last element from each list, you can use the following code:

interim = []
for x in tracked_output_sheet:interim.append(x[:-1])
tracked_output_sheet= interim