Mailing List Archive

Worked for me.

>>> l1=[5,3,7]
>>> l2=[8,6,3]
>>> x=[]
>>> for i in range(len(l1)):
... x.append(l1[i]+l2[i])
...
>>> x
[13, 9, 10]

--
--Darrell

Thorsten Zöller <tzoeller@t-online.de> wrote in message
news:7p1v28$vcf$1@news04.btx.dtag.de...
> Suppose you write a code as follows:
>
> x = [] # empty list
> list1 = [5, 3, 7] # list containing any
> numbers
> list2 = [8, 6, 3] # another list
containing
> any numbers
> for i in range(len(list1)):
> x.append(list1[i-1] + list2[i-1]) # [5+8, 3+6, 7+3]
> ...
> x # print resulting
> list
>
> I'm trying to achieve that, having two lists of the same lenght containig
> any numbers, each component of the two lists is added to his pendant of
the
> other lists (i.e. the first component of list1 is added to the first
> component of list2 etc). The result is saved in another list x. But if I
run
> this code, I do not get [13, 9, 19] for x, but [10, 13, 9], i.e. the
> components are in a wrong order now. Why does the order change and how can
I
> avoid it?
>
> Thank you
> Thorsten
>
>

"Thorsten Zöller" <tzoeller@t-online.de> writes:
| Suppose you write a code as follows:
| x = [] # empty list
| list1 = [5, 3, 7] # list containing any
| numbers
| list2 = [8, 6, 3] # another list containing
| any numbers
| for i in range(len(list1)):
| x.append(list1[i-1] + list2[i-1]) # [5+8, 3+6, 7+3]
| ...
| x # print resulting
| list
|
| I'm trying to achieve that, having two lists of the same lenght containig
| any numbers, each component of the two lists is added to his pendant of the
| other lists (i.e. the first component of list1 is added to the first
| component of list2 etc). The result is saved in another list x. But if I run
| this code, I do not get [13, 9, 19] for x, but [10, 13, 9], i.e. the
| components are in a wrong order now. Why does the order change and how can I
| avoid it?

Because you index the input lists with i-1. Use list1[i] + list2[i],
and then I think it will work as you want.

range() naturally returns a list that starts with 0, which happens to be
where Python sequences start as well. When you decrement that initial 0
index, -1 is like len(list1)-1, hence the last shall be first.

Donn Cave, University Computing Services, University of Washington
donn@u.washington.edu

Or even:

>>> import operator
>>> l1=[5,3,7]
>>> l2=[8,6,3]
>>> map( operator.add, l1,l2)
[13, 9, 10]
>>>

Enjoy,
Mike

-----Original Message-----
From: python-list-request@cwi.nl [mailto:python-list-request@cwi.nl]On
Behalf Of Darrell
Sent: August 13, 1999 4:59 PM
To: python-list@cwi.nl
Subject: Re: Question about order in lists


Worked for me.

>>> l1=[5,3,7]
>>> l2=[8,6,3]
>>> x=[]
>>> for i in range(len(l1)):
... x.append(l1[i]+l2[i])
...
>>> x
[13, 9, 10]

--
--Darrell

Thorsten Zöller <tzoeller@t-online.de> wrote in message
news:7p1v28$vcf$1@news04.btx.dtag.de...
> Suppose you write a code as follows:
>
> x = [] # empty list
> list1 = [5, 3, 7] # list containing any
> numbers
> list2 = [8, 6, 3] # another list
containing
> any numbers
> for i in range(len(list1)):
> x.append(list1[i-1] + list2[i-1]) # [5+8, 3+6, 7+3]
> ...
> x # print resulting
> list
>
> I'm trying to achieve that, having two lists of the same lenght containig
> any numbers, each component of the two lists is added to his pendant of
the
> other lists (i.e. the first component of list1 is added to the first
> component of list2 etc). The result is saved in another list x. But if I
run
> this code, I do not get [13, 9, 19] for x, but [10, 13, 9], i.e. the
> components are in a wrong order now. Why does the order change and how can
I
> avoid it?
>
> Thank you
> Thorsten
>
>

Thorsten Z÷ller asks:

> Suppose you write a code as follows:
>
> x = [] # empty list
> list1 = [5, 3, 7] # list containing
> any numbers list2 = [8, 6, 3] #
> another list containing any numbers for i in range(len(list1)):
> x.append(list1[i-1] + list2[i-1]) # [5+8, 3+6, 7+3]
> ...
> x # print
> resulting list
>
> I'm trying to achieve that, having two lists of the same lenght
> containig any numbers, each component of the two lists is added to
> his pendant of the other lists (i.e. the first component of list1 is
> added to the first component of list2 etc). The result is saved in
> another list x. But if I run this code, I do not get [13, 9, 19] for
> x, but [10, 13, 9], i.e. the components are in a wrong order now.
> Why does the order change and how can I avoid it?

Interactive is your friend!

>>> x = []
>>> list1 = [5,3,7]
>>> list2 = [8,6,3]
>>> for i in range(len(list1)):
... print "i is", i
... print "i-1 is", i-1
... print "list1[i-1] is", list1[i-1]
... print "list2[i-1] is", list2[i-1]
... x.append(list1[i] + list2[i]) #oops, that's the right way!
...
i is 0
i-1 is -1
list1[i-1] is 7
list2[i-1] is 3
i is 1
i-1 is 0
list1[i-1] is 5
list2[i-1] is 8
i is 2
i-1 is 1
list1[i-1] is 3
list2[i-1] is 6
>>> x
[13, 9, 10]
>>>

- Gordon

You win.
Map is very fast in this case.
Although something intresting happend when I used
l1[:] to create l2
##################
import operator, time

def mapIt(l1, l2):
map( operator.add, l1,l2)


def loopIt(l1, l2):
x=[]
append=x.append
for i in range(len(l1)):
append(l1[i]+l2[i])


sz=1000000
l1=[]
for i in xrange(sz):l1.append(i)

if 0:
# Time1: 5.75 sec
# Time2: 27.70 sec
l2=[]
for i in xrange(sz):l2.append(i)
else:
# Odd when using these huge numbers
# This way of creating l2 ran faster ??
# Time1: 3.25 sec
# Time2: 28.50 sec
l2=l1[:]

t1=time.time()
mapIt(l1,l2)
print " Time1: %.2f sec"%(time.time()-t1)
t1=time.time()
loopIt(l1,l2)
print " Time2: %.2f sec"%(time.time()-t1)

####################
# Time1: 3.25 sec
# Time2: 28.50 sec



--Darrell

Understandable behaviour:

l2 = l1[:] -> make a new list with size == l1 with pointers all pointing to
l1's data.

At the C level, this would mean approximately "copy memory range x to new
memory range y" (lists are contiguous arrays of pointers to objects).

Whereas:

for i in xrange(sz):l2.append(i)

Actually goes through x python loop iterations, where it binds the variable
i x times, finds attribute __getitem__ of your xrange x times, calls it x
times, finds the attribute append of l2 x times, and calls that function x
times. (And every single variable lookup has the potential to fall through
to a higher-level namespace etceteras). Python's dynamism has a _serious_
impact on performance and algo design ;) .

Oh, and the following is a fairly fast way to create your initial list of
integers (in this case, just a little faster):

range( 1000000 )

Cheers,
Mike

-----Original Message-----
From: python-list-request@cwi.nl [mailto:python-list-request@cwi.nl]On
Behalf Of Darrell
Sent: August 13, 1999 7:06 PM
To: python-list@cwi.nl
Subject: Re: Question about order in lists


You win.
Map is very fast in this case.
Although something intresting happend when I used
l1[:] to create l2
##################
import operator, time

def mapIt(l1, l2):
map( operator.add, l1,l2)


def loopIt(l1, l2):
x=[]
append=x.append
for i in range(len(l1)):
append(l1[i]+l2[i])


sz=1000000
l1=[]
for i in xrange(sz):l1.append(i)

if 0:
# Time1: 5.75 sec
# Time2: 27.70 sec
l2=[]
for i in xrange(sz):l2.append(i)
else:
# Odd when using these huge numbers
# This way of creating l2 ran faster ??
# Time1: 3.25 sec
# Time2: 28.50 sec
l2=l1[:]

t1=time.time()
mapIt(l1,l2)
print " Time1: %.2f sec"%(time.time()-t1)
t1=time.time()
loopIt(l1,l2)
print " Time2: %.2f sec"%(time.time()-t1)

####################
# Time1: 3.25 sec
# Time2: 28.50 sec



--Darrell