CODEWITHSUNDEEP
pythonstringlistconstructor
perror
perror

Reputation: 7386

What is the difference between list(a) and [a]?

I noticed a strange difference between two list constructors that I believed to be equivalent.

Here is a small example:

hello = 'Hello World'

first = list(hello)
second = [hello]

print(first)
print(second)

This code will produce the following output:

['H', 'e', 'l', 'l', 'o', ' ', 'W', 'o', 'r', 'l', 'd']
['Hello World']

So, the difference is quite clear between the two constructors... And, I guess that this could be generalized to other constructors as well, but I fail to understand the logic behind it.

Can somebody cast its lights upon my interrogations?

Upvotes: 1

Views: 256

Answers (6)

mementum
mementum

Reputation: 3203

The most important distinction of these 2 behaviours comes when you work with generators. Given that Python 3 transformed things like map and zip into generators ...

If we assume map returns generators:

a = list(map(lambda x: str(x), [1, 2, 3]))
print(a)

The result is:

['1', '2', '3']

But if we do:

a = [map(lambda x: str(x), [1, 2, 3])]
print(a)

The result is:

[<map object at 0x00000209231CB2E8>]

It is obvious that the 2nd case is in most situations undesirable and not expected.

P.S. If you are in Python 2, then do at the beginning: from itertools import imap as map

Upvotes: 1

chepner
chepner

Reputation: 530960

You are assuming that list(hello) should create a list containing one element, the object referred to by hello. That's not true; by that logic you would expect list(5) to return [5]. list takes a single iterable argument (a list, a tuple, a string, a dict, etc) and returns a list whose elements are taken from the given iterable.

The bracket notation, however, is not limited to containing a single item. Each comma-separated object is treated as a distinct element for the new list.

Upvotes: 1

Stefano
Stefano

Reputation: 3215

The first just transform the list "Hello world" (an character array) into a list 

first = list(hello)

The second create a list with element inside brackets.

first = [hello]

In the second case for example you could also do:

first = [hello, 'hi', 'world']

and as output of the print you will get

['Hello World', 'hi', 'world']

Upvotes: 2

deceze
deceze

Reputation: 522016

The list() constructor function takes exactly one argument, which must be an iterable. It returns a new list with each element being an element from the given iterable. Since strings are iterable (by character), a list with individual characters is returned.

[] takes as many "arguments" as you like, each being a single element in the list; the items are not "evaluated" or iterated, they are taken as is.

Everything as documented.

Upvotes: 8

Greg Hilston
Greg Hilston

Reputation: 2424

your "first" uses the list method, which takes in hello and treats it as an iterable, converting it to a list. Which is why each chararcter is seperate.

your "second" creates a new list, using the string as its value

Upvotes: 1

geo_pythoncl
geo_pythoncl

Reputation: 947

first = list(hello)

converts a string into a list.

second = [hello]

this places an item into a new list. it is not a constructor

Upvotes: 0

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