Fastest way to initialize a List in Python? (comparison of 3 options)

Fastest way to initialize a List in Python? (comparison of 3 options)
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#Python

[@GN69420]

Using [[0] * n] * n] creates a list of a list at the same point in memory, i.e. myList[0] is myList[1]

[@kimtgf7152]

Same with strings, I struggled so much while doing a programming question because of this

[@MaxMustermann-on2gd]

@@kimtgf7152 can expand on this? I wonder what the problem with strings is.

[@kimtgf7152]

@@MaxMustermann-on2gd Well, I had a n queens problem, and I had to make the board with each row being a string and all those strings were stored in a list. If I used something like ["...."]*n then what would happen is, if I changed one string all the other strings got changed as well.

[@Eknoma]

​@@kimtgf7152That shouldn't happen... how are you changing your strings? Python strings are immutable, so unless you changed the memory directly, that should be impossible

[@kimtgf7152]

@@Eknoma So I just went back and saw my code, I think I was just using strings in lists in list, because of that error was appearing, it had nothing to do with the strings but the lists of strings in list. I was later using join to convert it each row of list of strings into a string.

[@calum.macleod]

I love how Python optimises some things, in ways that we might not have expected. I really like your channel, I've only been programming Python for a year but you are helping to set important ideas in my head that were not obvious before. Thank you.

[@patloeber]

Happy to hear that!

[@pattmehta5835]

+1 for the disclaimer to not use this technique in a nested list. Very useful for scenarios where latency is critical

[@alexpaciniat]

Mutables can create tricky situations, whether as multiplicative elements in lists, where it is effectively the same object in memory and repeated, or when used as default argument in a function, where if modified it will effectively change the default value of the next call to that function.

[@GyroCannon]

The optimization is (assuming that the underlying structure is an array list) probably due to the code knowing how big to make the list. The other ways probably default to 10 or something and double in size each time it gets too big

[@shivanshmathur1287]

#Fastest as it uses numpy, which is a low level language import numpy as np n = 5 # Replace 5 with your desired size Listofzeros = np.zeros(n) MyList = list(Listofzeros) print(MyList)

[@pzkpfw917]

Sir, Loeber!! Thank you for the content on your YT channel.

[@ytusr-ko5rg]

I've recognized it when creating a 2d list: board1 = [[0]*3]*3 # would be different from board2 = [[0 for j in range(3)] for i in range(3)] # let's take a closer look # when I want to update the cell data of the board[row][column] >>> board1[0][0] = 1 >>> board1 [[1,0,0],[1,0,0],[1,0,0]] >>> board2[0][0] = 1 >>> board2 [[1,0,0],[0,0,0],[0,0,0]] # it's dangerous. Be careful with this python behavior:) # Using list-comprehension is the correct way.

[@bekhzodravshanov2161]

You cant multiply nested list. You have to use loop for creating it

[@MohammadBahiraei]

Wonderful

[@amrosamier8547]

Thank you..what IDE are you using please

[@devnarula6733]

looks like vscode to me

[@syukcode]

How to create Pointers in python?

[@gsravan7058]

Hi

[@MrJonas1995]

Why do you write the "_" in "for _ in list"? You usually use something like "for element in list", don't you?

[@MatrixAbuz17]

The ‘_’ in for loops indicates to the programmer that the temporary variable will be ignored.

[@Dekxterr]

what ide is this?

[@matinsasan]

Vs code

[@Dekxterr]

@@matinsasan Thankyou and cool pfp man