如何根据索引零 [0] 处的元组元素对 python 列表进行排序

Question

I have a python list right here

my_list = [("ww","hello"),("www","world"),("w","sardines")]

So I wanted to sort this list based on the length value of the first index of the tuple element which is as you can see there is the letter "w" . To be more specific the output should be like this

my_list = [("w","sardines"),("ww","hello"),("www","world")]

I've searched many sorting techniques for this but I find those with for example a string element only. I was thinking of using a sorting algorithm for this like bubble sort, merge sort, insertion or any other sorting algorithm... but I think I am lost in that part and also I think that there is a better way to do this without even using a sorting algorithm. I feel like there's a simpler way like just calling a .sort() function.

Thank you so much for your help! Glad to ask questions here in this great community! Thank you!

Answer 1

Try this:

my_list = [("ww","hello"),("www","world"),("w","sardines")]

def take_el(elem):
    return len(elem[0])

sorted(my_list, key=take_el)

or with lambda

sorted(my_list, key = lambda x: len(x[0]))

Answer 2

Use this simple technique to sort on the basis of 1st element in tuple:

sorted(my_list, key=lambda x: len(x[0]))

Change x[0] to x[1] to sort the list using the length of 2nd tuple element and so on..

Answer 3

my_list = [("ww","hello"),("www","world"),("w","sardines")]
listOrder = []
newList = []
for i in my_list:
    listOrder.append(int(len(i[0])))
listOrder.sort()
for i in listOrder:
    for j in my_list:
        if i == len(j[0]):
            newList.append(j)

print(newList)

This will also help.

Answer 4

use sorted where x is the tuple and x[n] is the slice

my_list = [("ww","hello"),("www","world"),("w","sardines")]

result = sorted(my_list, key = lambda x: x[0])
print(result)