A little more advanced for loop

[Horta]

Hi folks,

Suppose I have to loop over 3 lists being the same size at the same
time and order. How can I do that without using the range() function
or whatever indexing?

Example using range:

a = ['aaa', 'aaaa']
b = ['bb', 'bbbb']
c = ['c', 'cccc']

for i in range(len(a)):
# using a, b, and c

I'm sure there's a elegant way to do that...

Thanks in advance.

 

[Diez B. Roggisch]

Hi folks,

Suppose I have to loop over 3 lists being the same size at the same
time and order. How can I do that without using the range() function
or whatever indexing?

Example using range:

a = ['aaa', 'aaaa']
b = ['bb', 'bbbb']
c = ['c', 'cccc']

for i in range(len(a)):
# using a, b, and c

I'm sure there's a elegant way to do that...

Use zip:

for av, bv, cv in zip(a, b, c):
print av, bv, cv

Diez

 

[Stephan Diehl]

Hi folks,

Suppose I have to loop over 3 lists being the same size at the same
time and order. How can I do that without using the range() function
or whatever indexing?

Example using range:

a = ['aaa', 'aaaa']
b = ['bb', 'bbbb']
c = ['c', 'cccc']

for i in range(len(a)):
# using a, b, and c

I'm sure there's a elegant way to do that...

Thanks in advance.

Sure, there is:

for a_item, b_item , c_item in zip(a,b,c):
# do something

 

[Horta]

Hi folks,

Suppose I have to loop over 3 lists being the same size at the same
time and order. How can I do that without using the range() function
or whatever indexing?

Example using range:

a = ['aaa', 'aaaa']
b = ['bb', 'bbbb']
c = ['c', 'cccc']

for i in range(len(a)):
# using a, b, and c

I'm sure there's a elegant way to do that...

Thanks in advance.

Sure, there is:

for a_item, b_item , c_item in zip(a,b,c):
# do something

Thanks guys!

 

[Larry Bates]

Hi folks,
Suppose I have to loop over 3 lists being the same size at the same
time and order. How can I do that without using the range() function
or whatever indexing?
Example using range:
a = ['aaa', 'aaaa']
b = ['bb', 'bbbb']
c = ['c', 'cccc']
for i in range(len(a)):
# using a, b, and c
I'm sure there's a elegant way to do that...
Thanks in advance.

Sure, there is:

for a_item, b_item , c_item in zip(a,b,c):
# do something

Thanks guys!

Note: if lists are long take a look at itertools izip. zip creates
a list of lists which could take lots of memory/time if they are VERY
large. itertools izip iterates over them in place.

from itertools import izip

for a_item, b_item, c_item in izip(a,b,c):
# do something

-Larry

 

[Duncan Booth]

Note: if lists are long take a look at itertools izip. zip creates
a list of lists which could take lots of memory/time if they are VERY
large. itertools izip iterates over them in place.

That's interesting. I was going to quibble with the assertion that zip
could take lots of time, since on the face of it a loop using izip packs
and unpacks just as many tuples. Fortunately I tried it out before claiming
that zip would be just as fast

It would appear that even for short sequences the izip solution is faster.
My guess would be it is because the same tuple objects are being reused in
the izip version.

C:\Python25\Lib>timeit.py -s "a, b, c = [range(1000)]*3" -s "from itertools
import izip" "for p,q,r in izip(a,b,c): pass"
10000 loops, best of 3: 131 usec per loop

C:\Python25\Lib>timeit.py -s "a, b, c = [range(1000)]*3" "for p,q,r in
zip(a,b,c): pass"
1000 loops, best of 3: 212 usec per loop

C:\Python25\Lib>timeit.py -s "a, b, c = [range(100)]*3" -s "from itertools
import izip" "for p,q,r in izip(a,b,c): pass"

100000 loops, best of 3: 13.9 usec per loop

C:\Python25\Lib>timeit.py -s "a, b, c = [range(100)]*3" "for p,q,r in
zip(a,b,c): pass"
10000 loops, best of 3: 22.6 usec per loop

C:\Python25\Lib>timeit.py -s "a, b, c = [range(10)]*3" -s "from itertools
import izip" "for p,q,r in izip(a,b,c): pass"
100000 loops, best of 3: 2.21 usec per loop

C:\Python25\Lib>timeit.py -s "a, b, c = [range(10)]*3" "for p,q,r in
zip(a,b,c): pass"
100000 loops, best of 3: 3.52 usec per loop

 

[Aahz]

Note: if lists are long take a look at itertools izip. zip creates
a list of lists which could take lots of memory/time if they are VERY
large. itertools izip iterates over them in place.

That's half-true -- while izip is faster, it's not enough faster to make
a significant difference unless the lists are "huge" rather than
"large". Remember that the data elements themselves are not copied.