CODEWITHSUNDEEP
pythonconstructor
abalter
abalter

Reputation: 10383

Strangeness with type constructors in python

In python I can do things like:

d = dict()
i = int()
f = float()
l = list()

but there is no constructor for strings

>>> s = string()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'string' is not defined

Why is that? Or, IS there a constructor for string types?

Also, following the definitions above,

d['a'] = 1

works, but

>>> l[0] = 1
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: list assignment index out of range

does not.

Can someone explain this to me?

Upvotes: 1

Views: 109

Answers (4)

mgilson
mgilson

Reputation: 310019

There is a builtin function for string creation:

s = str()  # s == ''

As far as your assignment into a list vs your assignment into a dictionary:

Dictionaries are unordered, so it makes complete sense to be able to add a key value pair anywhere at any time. In fact, dictionary keys don't even need to be the same type (d[1]='foo'; d["string"]='bar'). Lists however are ordered. Consider the following:

a=list()
a[1]='foo'

What should the language do in that scenario? a[0] hasn't yet been defined, so the language would have to make something up in order to maintain the ordered-ness of lists, or throw an error. Assigning to a[0] or the next element in the sequence (e.g. a=list(); a[0]='bar') is a very special case and from reading the "Zen of Python", special cases aren't special enough to break the rules. I would guess this is why Guido decided to force the list element to exist before you can assign to it (e.g a=list(); a.append('foo'); a[0]='bar')

Upvotes: 1

Alex Wilson
Alex Wilson

Reputation: 6740

You meant:

a = str()

I believe. Then everything works.

And as to the distinction between d['a'] and l[0]: d is a dictionary, which has a sparse representation of the elements stored. Whereas a list (l) represents data densely: so if you had:

l = [1,2]

And you subsequently wrote:

l[200] = 31

It would imply, as well as assigning to element 200, to be entirely consistent, putting some arbitrary values in l[3:199] as well.

Upvotes: 3

senderle
senderle

Reputation: 151047

As others have noted, you want str, not string.

But to answer your other question, lists cannot be extended by assignment. If you try to assign outside the bounds of a list, you get an error. On the other hand, dictionaries have no bounds in any meaningful sense; they have only defined keys and undefined keys.

Think of a dictionary as a bag of objects tied together, and a list as a tray with a fixed number of compartments. You can throw a pair of things into the bag anytime, but you can't put something in a tray's compartment if no such compartment exists. You have to create the compartment, using append or something similar. Since your "tray" has no compartments yet, l[0] = x fails.

Upvotes: 2

Ashwini Chaudhary
Ashwini Chaudhary

Reputation: 251021

yes we've str():

>>> s=str()
>>> s
''

Upvotes: 1

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