Print a list in reverse order with range()?

Use the 'range' built-in function. The signature is range(start, stop, step). This produces a sequence that yields numbers, starting with start, and ending if stop has been reached, excluding stop.

>>> range(9,-1,-1) range(9, -1, -1) >>> list(range(9,-1,-1)) [9, 8, 7, 6, 5, 4, 3, 2, 1, 0] >>> list(range(-2, 6, 2)) [-2, 0, 2, 4] 

The list constructor converts range (which is a python generator), into a list.

You could use range(10)[::-1] which is the same thing as range(9, -1, -1) and arguably more readable (if you're familiar with the common sequence[::-1] Python idiom).

For those who are interested in the "efficiency" of the options collected so far...

Jaime RGP's answer led me to restart my computer after timing the somewhat "challenging" solution of Jason literally following my own suggestion (via comment). To spare the curious of you the downtime, I present here my results (worst-first):^[1]

Jason's answer (maybe just an excursion into the power of list comprehension):

$ python -m timeit "[9-i for i in range(10)]" 1000000 loops, best of 3: 1.54 usec per loop 

martineau's answer (readable if you are familiar with the extended slices syntax):

$ python -m timeit "range(10)[::-1]" 1000000 loops, best of 3: 0.743 usec per loop 

Michał Šrajer's answer (the accepted one, very readable):

$ python -m timeit "reversed(range(10))" 1000000 loops, best of 3: 0.538 usec per loop 

bene's answer (the very first, but very sketchy at that time):

$ python -m timeit "range(9,-1,-1)" 1000000 loops, best of 3: 0.401 usec per loop 

The last option is easy to remember using the range(n-1,-1,-1) notation by Val Neekman.

^[1] As commented by Karl Knechtel, the results presented here are version-dependent and refer to the Python 3.x version that was stable at the time of answering this question.

No sense to use reverse because the range function can return reversed list.

When you have iteration over n items and want to replace order of list returned by range(start, stop, step) you have to use third parameter of range which identifies step and set it to -1, other parameters shall be adjusted accordingly:

1. Provide stop parameter as -1 (it's previous value of stop - 1, stop was equal to 0).
2. As start parameter use n-1.

So equivalent of range(n) in reverse order would be:

n = 10 print range(n-1,-1,-1) #[9, 8, 7, 6, 5, 4, 3, 2, 1, 0] 

for i in range(8, 0, -1) 

will solve this problem. It will output 8 to 1, and -1 means a reversed list

Very often asked question is whether range(9, -1, -1) better than reversed(range(10)) in Python 3? People who have worked in other languages with iterators immediately tend to think that reversed() must cache all values and then return in reverse order. Thing is that Python's reversed() operator doesn't work if the object is just an iterator. The object must have one of below two for reversed() to work:

1. Either support len() and integer indexes via []
2. Or have __reversed__() method implemented.

If you try to use reversed() on object that has none of above then you will get:

>>> [reversed((x for x in range(10)))] TypeError: 'generator' object is not reversible 

So in short, Python's reversed() is only meant on array like objects and so it should have same performance as forward iteration.

But what about range()? Isn't that a generator? In Python 3 it is generator but wrapped in a class that implements both of above. So range(100000) doesn't take up lot of memory but it still supports efficient indexing and reversing.

So in summary, you can use reversed(range(10)) without any hit on performance.

Readibility aside, reversed(range(n)) seems to be faster than range(n)[::-1].

$ python -m timeit "reversed(range(1000000000))" 1000000 loops, best of 3: 0.598 usec per loop $ python -m timeit "range(1000000000)[::-1]" 1000000 loops, best of 3: 0.945 usec per loop 

Just if anyone was wondering :)

The requirement in this question calls for a list of integers of size 10 in descending order. So, let's produce a list in python.

# This meets the requirement. # But it is a bit harder to wrap one's head around this. right? >>> range(10-1, -1, -1) [9, 8, 7, 6, 5, 4, 3, 2, 1, 0] # let's find something that is a bit more self-explanatory. Sounds good? # ---------------------------------------------------- # This returns a list in ascending order. # Opposite of what the requirement called for. >>> range(10) [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] # This returns an iterator in descending order. # Doesn't meet the requirement as it is not a list. >>> reversed(range(10)) <listreverseiterator object at 0x10e14e090> # This returns a list in descending order and meets the requirement >>> list(reversed(range(10))) [9, 8, 7, 6, 5, 4, 3, 2, 1, 0] 

You can do printing of reverse numbers with range() BIF Like ,

for number in range ( 10 , 0 , -1 ) : print ( number ) 

Output will be [10,9,8,7,6,5,4,3,2,1]

range() - range ( start , end , increment/decrement ) where start is inclusive , end is exclusive and increment can be any numbers and behaves like step

i believe this can help,

range(5)[::-1] 

below is Usage:

for i in range(5)[::-1]: print i 

Using without [::-1] or reversed -

def reverse(text): result = [] for index in range(len(text)-1,-1,-1): c = text[index] result.append(c) return ''.join(result) print reverse("python!") 

range(9,-1,-1) [9, 8, 7, 6, 5, 4, 3, 2, 1, 0] 

I thought that many (as myself) could be more interested in a common case of traversing an existing list in reversed order instead, as it's stated in the title, rather than just generating indices for such traversal.

Even though, all the right answers are still perfectly fine for this case, I want to point out that the performance comparison done in Wolf's answer is for generating indices only. So I've made similar benchmark for traversing an existing list in reversed order.

TL;DR a[::-1] is the fastest.

NB: If you want more detailed analysis of different reversal alternatives and their performance, check out this great answer.

Prerequisites:

a = list(range(10)) 

Jason's answer:

%timeit [a[9-i] for i in range(10)] 1.27 µs ± 61.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) 

martineau's answer:

%timeit a[::-1] 135 ns ± 4.07 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each) 

Michał Šrajer's answer:

%timeit list(reversed(a)) 374 ns ± 9.87 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) 

bene's answer:

%timeit [a[i] for i in range(9, -1, -1)] 1.09 µs ± 11.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) 

As you see, in this case there's no need to explicitly generate indices, so the fastest method is the one that makes less extra actions.

NB: I tested in JupyterLab which has handy "magic command" %timeit. It uses standard timeit.timeit under the hood. Tested for Python 3.7.3

[9-i for i in range(10)] [9, 8, 7, 6, 5, 4, 3, 2, 1, 0] 

because range(n) produces an iterable there are all sorts of nice things you can do which will produce the result you desire, such as:

range(n)[::-1] 

if loops are ok, we can make sort of a queue:

a = [] for i in range(n): a.insert(0,a) return a 

or maybe use the reverse() method on it:

reverse(range(n)) 

You don't necessarily need to use the range function, you can simply do list[::-1] which should return the list in reversed order swiftly, without using any additions.

Suppose you have a list call it a={1,2,3,4,5} Now if you want to print the list in reverse then simply use the following code.

a.reverse for i in a: print(i) 

I know you asked using range but its already answered.

You could use a User-Defined Function like this one:

def range_reversed(start: int, stop: int, step: int) -> range: # Determine the number of elements in the sequence num_elem = len(range(start, stop, step)) # Calculate the start of the reverse sequence using the formula for an arithmetic sequence start_reversed = start + (num_elem - 1) * step # Determine the stop of the reverse sequence stop_reversed = start - 1 if step > 0 else start + 1 return range(start_reversed, stop_reversed, -step) 

# Example usage: print(list(range_reversed(0,13,4))) # Output: [12, 8, 4, 0] 

range(9,-1,-1) [9, 8, 7, 6, 5, 4, 3, 2, 1, 0] 

Is the correct form. If you use

reversed(range(10)) 

you wont get a 0 case. For instance, say your 10 isn't a magic number and a variable you're using to lookup start from reverse. If your n case is 0, reversed(range(0)) will not execute which is wrong if you by chance have a single object in the zero index.