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I'm trying to make a list with numbers 1-1000 in it. Obviously this would be annoying to write/read, so I'm attempting to make a list with a range in it. In Python 2 it seems that:

some_list = range(1,1000)

would have worked, but in Python 3 the range is similar to the xrange of Python 2?

Can anyone provide some insight into this?

You can just construct a list from the range object:

my_list = list(range(1, 1001))

This is how you do it with generators in python2.x as well. Typically speaking, you probably don't need a list though since you can come by the value of my_list[i] more efficiently (i + 1), and if you just need to iterate over it, you can just fall back on range.

Also note that on python2.x, xrange is still indexable1. This means that range on python3.x also has the same property2

1print xrange(30)[12] works for python2.x

2The analogous statement to 1 in python3.x is print(range(30)[12]) and that works also.

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In Pythons <= 3.4 you can, as others suggested, use list(range(10)) in order to make a list out of a range (In general, any iterable).

Another alternative, introduced in Python 3.5 with its unpacking generalizations, is by using * in a list literal []:

>>> r = range(10)
>>> l = [*r]
>>> print(l)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Though this is equivalent to list(r), it's literal syntax and the fact that no function call is involved does let it execute faster. It's also less characters, if you need to code golf :-)

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in Python 3.x, the range() function got its own type. so in this case you must use iterator

list(range(1000))

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You really shouldn't need to use the numbers 1-1000 in a list. But if for some reason you really do need these numbers, then you could do:

[i for i in range(1, 1001)]

List Comprehension in a nutshell:

The above list comprehension translates to:

nums = []
for i in range(1, 1001):
    nums.append(i)

This is just the list comprehension syntax, though from 2.x. I know that this will work in python 3, but am not sure if there is an upgraded syntax as well

Range starts inclusive of the first parameter; but ends Up To, Not Including the second Parameter (when supplied 2 parameters; if the first parameter is left off, it'll start at '0')

range(start, end+1)
[start, start+1, .., end]

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Does range() return a list or a range object?, In Python 3, range stopped generating lists and became essentially what xrange One way is to turn the ranges into lists before concatenation: To add another alternative to tuple(l), as of Python >= 3.5 you can do: t = *l, # or t = (*l,) short, a bit faster but probably suffers from readability. This essentially unpacks the list l inside a tuple literal which is created due to the presence of the single comma ,.

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Comments
  • also, some_list[i] == i+1, so you probably don't really need a list anyway.
  • @RikPoggi. for example, one might need to supply a list for a plotting function. Sometimes a range will suffice, but a range cannot be concatenated (is immutable), so if you need to add a default starting value to all lists being plotted, that on needs to be turned into a list also.
  • This is definitely the way to go, but a nitpick: this isn't really a "cast"
  • @jterrace changed "cast" to "convert". You're right about it not being a cast... I don't really know what to call it exactly.
  • I would say "construct" or "build" (or possibly "materialise")- as you're not "converting" (as such) a generator to a list, you're creating a new list object from a data source which happens to be a generator... (but s'pose just splitting hairs and not 100% sure what I favour anyway)
  • My +1 for "construct" as it is consistent with other OO languages. The list(arg) is understood in other languages as calling a constructor of the list class. Actually, it is also the Python case. The debates whether the object is filled during the construction (as in the C++ case) or only during the first automatically called method (as in the Python __init__() method) cannot change the basic abstract idea. My view is that the list constructor takes the iterator and fills the list with the returned values.
  • Why does it give an error in jupyter notebook and working fine in shell? Error: 'range' object is not callable
  • To be clear, you can still one-line it: [*range(10)] works just fine for when you don't need the range for any purpose but initializing the list. Side-note: My favorite(okay, not really) part of the unpacking generalizations is that empty sets now have a literal syntax, {*()}, or as I call it, the one-eyed monkey operator. ;-)
  • @ShadowRanger that's how I originally thought about writing it. I decided to be a bit more verbose in order to not confuse new Python users :-)
  • "in this case you must use iterator"? What the heck is that supposed to mean?
  • I am using python 3.7 & tried x=list(range(1000)) but got the error TypeError: 'list' object is not callable
  • @Earthshaker You must have a typo, like list(range(1000))()