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Iterating over a Python generator


By : Steve P.
Date : September 15 2020, 10:00 PM
hop of those help? I'm trying to write a function that will take as input a string of intervals e.g "0-0,4-8,20-21,43-45" to produce all numbers within each of the ranges meaning: [0, 4, 5, 6, 7, 8, 20, 21, 43, 44, 45]
code :
def parse_ranges(arg):
    arg = arg.split(",")
    parsed = ((line.split("-")) for line in arg)

    for pair in parsed:
        yield from range(int(pair[0]), int(pair[1])+1)

print(list(parse_ranges("0-0,4-8,20-21,43-45")))

Out: [0, 4, 5, 6, 7, 8, 20, 21, 43, 44, 45]


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Python nested for loop Vs generator for repeatedly iterating over a list provided by a function


By : Jakorat
Date : March 29 2020, 07:55 AM
around this issue The problem is quite simple, I have a method that returns a list. I want to iterate over every item in this list, and once finished, call the method to receive a new list, and repeat. , I'd suggest you use some iterators
code :
items = itertools.chain.from_iterable( iter(list_returning_method, None) )

for item in items:
    # do something with item
    print item

    if not Condition:
       break

Python: why is it faster to iterate over a list than iterating over an xrange() generator define on its length?


By : JhayCo
Date : March 29 2020, 07:55 AM
I hope this helps you . I did it in Python3 , but the same results arose. I put the range creation in setup for a more accurate comparison
code :
In [1]: timeit.timeit('for _ in a: continue', setup='a=list(range(10000))', number=10000)
Out[1]: 1.195666481000444

In [2]: timeit.timeit('for _ in a: continue', setup='a=range(10000)', number=10000)
Out[2]: 2.4083170039994
In [3]: timeit.timeit('for _ in range(10000): continue', number=10000)
Out[3]: 4.166428555001403

In [4]: timeit.timeit('for _ in list(range(10000)): continue', number=10000)
Out[4]: 5.800707030000922

Python : generator object when iterating list


By : user3714200
Date : March 29 2020, 07:55 AM
I wish did fix the issue. You need to wrap the generator expression in square brackets to turn it into a list:
code :
c=[i for i in a]
c = a + b

python joblib.load: Iterating over generator results in infinite loop


By : Eduardo
Date : March 29 2020, 07:55 AM
I hope this helps . According to the documentation, joblib.load returns the full object.
code :
def load(filename):
    with open(filename, "rb") as f:
        return joblib.load(f)

Python multiple processes consuming/iterating over single generator (divide and conquer)


By : Mark Howard
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Most simple solution with least code? multiprocessing context manager.
I assume that you can put "do something with string" into a function called "do_something"
code :
from multiprocessing import Pool as ProcessPool

number_of_processes = 4

with ProcessPool(number_of_processes) as pool:
    pool.map(do_something, my_strings)
with ProcessPool(number_of_processes) as pool:
    results = pool.map(do_something, my_strings)
from multiprocessing.dummy import Pool as ThreadPool
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