来自带有 while 循环的输入的 Python 函数子列表

Question

My problem is this: Write a function, sublist, that takes in a list of numbers as the parameter. In the function, use a while loop to return a sublist of the input list. The sublist should contain the same values of the original list up until it reaches the number 5 (it should not contain the number 5).

I tinker with it and I get the question partially correct sometimes.

def sublist(x):
    a = [int(x) for x in input()]
    while x < 5:
        x = x + 1
    return(x)

Answer 1

Try:

import itertools

def sublist(x):
    return list(itertools.takewhile(lambda n: n != 5, x))

Update : If this is a homework question, my answer won't work for you - but nor should we just give you an answer , so, look at while and break . Think about creating an empty list to start with, adding things to it until you need to stop, then returning it.

Answer 2

If you want to use a while loop with a check on the number value, you'd better create a generator from the input list and use next() to iterate over it:

def sublist(x):
    sub = []
    x = (num for num in x)  # create a generator
    num = next(x, 5)  
    while num != 5:
        sub.append(num)
        num = next(x, 5)  # iterate
    return sub

x = [1, 3, 4, 5, 1, 2, 3]
sublist(x)

>>> [1, 3, 4]

Answer 3

  def sublist(lst):
    sub = lst
    for i in lst:
       if i == 5:
          x = lst.index(i)
          sub = lst[:x]
    return sub

  list3 = [1,56,34,2,3,1,4,5,5,5]
  sublist(list3) 

---

I want to give an answer with using slicing and index method. that's very useful for Python beginners.

Answer 4

If you are looking to write a function that takes every element of a list until the number 5 (eg [1, 2, 3, 4, 5] -> [1, 2, 3, 4]) then you could do that like this:

def sublist(input_list):
    output_list = []
    index = 0
    while index < len(input_list):
        if input_list[index] != 5:
            output_list.append(input_list[index])
            index += 1
        else:
            break
    return output_list

The while loop is broken when you reach 5. Until then, each value is added to a new list which is then returned by the function.

Update: Change while condition to check index is less than the length of the input list

Answer 5

It seems like you don't understand the question.

Write a function, sublist , that takes a list of numbers as the parameter.

This means that if we have this:

def sublist(x):
    pass

then x is going to be a list — not , as in your example, a number. Also, you don't need to do anything with input() ; you've already got the list, so don't need that line at all.

In the function, use a while loop to return a sublist of the input list.

Well, Python has a feature called "generators" that let you do this very easily, I'm going to cheat a bit, and not use a while loop. Instead, I'll use a for loop:

def sublist(x):
    for num in x:
        if num == 5:
            # we need to stop; break out of the for loop
            break
        # output the next number
        yield num

Now this code works:

>>> for num in sublist([3, 4, 2, 5, 6, 7]):
...     print(num)
3
4
2
>>> 

However, sublist doesn't technically return a list . Instead, let's use some MAGIC to make it return a list:

from functools import wraps
return_list = lambda f:wraps(f)(lambda *a,**k:list(f(*a,**k)))

(You don't need to know how this works.) Now, when we define our function, we decorate it with return_list , which will make the output a list :

@return_list
def sublist(x):
    for num in x:
        if num == 5:
            # we need to stop; break out of the for loop
            break
        # output the next number
        yield num

And now this also works:

>>> print(sublist([3, 4, 2, 5, 6, 7]))
[3, 4, 2]
>>>

Hooray!

Answer 6

def sublist(x):
    accum = 0
    sub = []
    while accum < len(x):
        if x[accum]== 5:
            return sub
        else:
            sub.append(x[accum])
        accum = accum +1
    return sub

x = [1, 3, 4, 5,6,7,8]
print(sublist(x))

Answer 7

def sublist(lst):


    output_list = []
    for i in lst:
        while i==5:
            return output_list
        output_list.append(i)
    return output_list

Answer 8

I did this like that.

def sublist(x):
    count = 0
    new = []
    if 5 in x:
        while(x[count] != 5):
            new.append(x[count])
            count += 1
        return new
        
    else:
        return x

Answer 9

def sublist(num):
    list = []
    for x in num:
        if x == 5:
            break
        while x != 5:
            list.append(x)
            break
    return list

num = [1,2,4,6,7,8,9,5,1,3,4]
print(sublist(num))

Answer 10

num = [1, 2, 3, 17, 1, 3, 5, 4, 3, 7, 5, 6, 9]
new = []
def check_nums(x):
    idx = 0
    while idx < len(x) and x[idx] != 7:
        print(idx)
        new.append(x[idx])
        idx += 1
    print(idx)
    return new
print(check_nums(num))

Answer 11

def sublist(a):
    i=0;
    lst=[];
    while(i<len(a) and a[i]!=5):
        lst.append(a[i]);
        i+=1;
    return lst;