list only stops once the element of the list is th

Question

I want to write a code that contains a sublist , that only stops once the element of the list is a certain number , for example 9.

I´ve already tried using different operators , if statements .

def sublist (list): 
return [x for x in list if x  <9]

[7,8,3,2,4,9,51] the output for the list above should be : [7,8,3,2,4]

Answer 1

To begin with, it's not a good idea to use python keywords like list as variable.

The list comprehension [x for x in list if x < 9] filters out elements less than 9, but it won't stop when it encounters a 9, instead it will go over the entire list

Example:

li = [7,8,3,2,4,9,51,8,7]
print([x for x in li if x < 9])

The output is

[7, 8, 3, 2, 4, 8, 7]

To achieve what you are looking for, you want a for loop which breaks when it encounters a given element (9 in your case)

li = [7,8,3,2,4,9,51]

res = []
item = 9

#Iterate over the list
for x in li:
    #If item is encountered, break the loop
    if x == item:
        break
    #Append item to list
    res.append(x)

print(res)

The output is

[7, 8, 3, 2, 4]

Answer 2

List comprehensions really are for making mapping/filtering combinations. If the length depends on some previous state in the iteration, you're better off with a for-loop, it will be more readable. However, this is a use-case for itertools.takewhile. Here is a functional approach to this task, just for fun, some may even consider it readable:

>>> from itertools import takewhile
>>> from functools import partial
>>> import operator as op
>>> list(takewhile(partial(op.ne, 9), [7,8,3,2,4,9,51]))
[7, 8, 3, 2, 4]

Answer 3

You can use iter() builtin with sentinel value (official doc)

l = [7,8,3,2,4,9,51]
sublist = [*iter(lambda i=iter(l): next(i), 9)]
print(sublist)

Prints:

[7, 8, 3, 2, 4]