Python: encontrar la posición del valor en una lista y generar una nueva lista

Currently learning python in my course and I'm a little confused on this part, unfortunately my next prac isn't until next week so I thought I'd ask here.

We're supposed to write a function called find() that takes a list, searches for the value in the list and returns as a new list of the positions in the list the number was found. Eg:

list = 2, 3, 7, 2, 3, 3, 5
number = 3

la salida sería: 0, 4, 5

The question requires us to use a loop but not to use built-in functions, slice expressions, list methods, or string methods unless specified, with the only ones specified for this part being: .append() y range().

We're given a file we're not allowed to edit:

import test_lists

list_A = ['q', 'w', 'e', 'r', 't', 'y']

print(test_lists.find(list_A, 'r'))
print(test_lists.find(list_A, 'b'))

From what I've attempted in a file named test_lists, the output gives <function find at 0x0000000003B89D90>:

def find(list1, listValue):

    findList = 0

    for x in range(findList):

        findList.find(list1, value)

    return find

anyone able to please push me in the right direction and explain it to me? We were told we could use .append() but I don't seem to understand how that would fit into this situation as from what I am aware append only adds to to the string. I feel like I'm pretty far off track.

Muchas Gracias

preguntado el 28 de mayo de 14 a las 13:05

Quieres decir 1, 4, 5 ? -

Hm, I assumed it would count up from 0 and not from 1. That was just an example anyway. Looking to find the position of r in that list in the part below -

4 Respuestas

Usar enumerate y lista de comprensión:

>>> lst = 2, 3, 7, 2, 3, 3, 5
>>> number = 3
>>> [i for i, x in enumerate(lst) if x == number]
[1, 4, 5]

>>> list_A = ['q', 'w', 'e', 'r', 't', 'y']
>>> [i for i, x in enumerate(list_A) if x == 'r']

Por cierto, no uses list as a variable name. It shadows builtin type/function list.


sin uso enumerate:

>>> lst = 2, 3, 7, 2, 3, 3, 5
>>> number = 3
>>> [i for i in range(len(lst)) if lst[i] == number]
[1, 4, 5]

contestado el 28 de mayo de 14 a las 13:05

thanks for the reply however as stated in my question we can't use built in functions which would include enumerate(). We can only use range() función y .append() method to get the output. - Asphyxiate9

Well then the quick change is: [i for i in range(len(lst)) if lst[i] == number] - RemolachaDemGuise

Will this do ?

def find(list1, listValue):
    found = []
    for index in range(0,len(list1)):
        if list1[index]==listValue:
    return found

contestado el 28 de mayo de 14 a las 13:05

I guess you can't use enumerate ?

Then, you you have to do the same thing by yourself : iterate over the indexes of the list and for each index, see if the element is the expected element. If it is, you had the index into your result list :

expected_element = 3
my_list = [...]
result_list = []
for i in range(len(my_list)):
    if my_list[i] == expected_element:

It's not very pythonic but if you have so much contraint, i think it's good

Y si no puedes usar len(my_list), you can get the size by iterate over a first time :

def size(l):
    s = 0
    for _ in l:
        s += 1
    return s

contestado el 28 de mayo de 14 a las 13:05

A list comprehension would be the most Pythonic way to do this. But writing it explicitly for instructional purposes, here is another way to do it.

def find(number, numList):
    indexMatches = []                    # initialize your index array
    for index in range(len(numList)):    # iterate over the list of values
        if number == numList[index]:     # compare the number against the current value
            indexMatches.append(index)   # if they match, add it to the index array
    return indexMatches                  # return the list of indexes where matches occured

find(3, [2,3,7,2,3,3,5])


[1, 4, 5]

contestado el 28 de mayo de 14 a las 13:05

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