CODEWITHSUNDEEP
python
DIVYANSHU
DIVYANSHU

Reputation: 13

No errors but output is unexpected

this is my code

for i in range(1,5): 
    print(i for x in range(1,i+1))

for this code the output is

<generator object at 0x0000025B9C7A9BA0>
<generator object at 0x0000025B9C7A9BA0>
<generator object at 0x0000025B9C7A9BA0>
<generator object at 0x0000025B9C7A9BA0>

Upvotes: 1

Views: 78

Answers (2)

Christoph Rackwitz
Christoph Rackwitz

Reputation: 15456

When you write

i for x in range(1,i+1)

that is a "generator expression" and it results in a "generator" (<generator object at ...>), which is an "iterable". Lists are also iterables. A generator's advantage is that it only actually evaluates the element expression (here i) when you iterate over it.

You seem to want an actual list though. You can immediately create a list like so:

[i for x in range(1,i+1)]

Or in your code:

for i in range(1,5): 
    print([i for x in range(1,i+1)])

The expression [... for ... in ...] is a "list comprehension". It looks similar to a generator expression, but it creates a list directly, without a generator. List comprehensions came first to Python, then came generator expressions. They have the same syntax inside different braces.

If you have a generator already, you could also turn it into a list:

for i in range(1,5): 
    gen = (i for x in range(1,i+1))
    print(list(gen)) # get all the elements and make a list

That generator is then "used up" and has no more elements to give.

Upvotes: 1

I&#39;mahdi
I&#39;mahdi

Reputation: 24059

you need create list.

try this:

for i in range(1,5): 
    print([i for x in range(1,i+1)])

in short format you can try this:

print([i for i in range(1,5) for x in range(1,i+1)])

or

list(map((lambda x : print([x for _ in range(1,x+1)])), range(1,5)))

or by thanks of @Sujay for run faster you can use tuple like below:

tuple(map((lambda x : print([x for _ in range(1,x+1)])), range(1,5)))

for check runtime i use below code:

%timeit tuple(map((lambda x : print([x for _ in range(1,x+1)])), range(1,5)))
# 794 µs ± 86.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit list(map((lambda x : print([x for _ in range(1,x+1)])), range(1,5)))
# 806 µs ± 150 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Upvotes: 4

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