Re: A bug in Partition?
- To: mathgroup at smc.vnet.net
- Subject: [mg117401] Re: A bug in Partition?
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Thu, 17 Mar 2011 06:32:53 -0500 (EST)
On 3/16/11 at 6:30 AM, lehin.p at gmail.com (Alexey) wrote:
>Consider the following:
>In[3]:=
>Partition[{a,b,c,d,e,f,g,i},3,3,-1,x]
>Partition[{a,b,c,d,e,f,g,i},3,3,1,x]
>Out[3]=
>{{x,x,a},{b,c,d},{e,f,g}}
>Out[4]=
>{{a,b,c},{d,e,f},{g,i,x}}
>One can see that in the first case element 'i' is dropped! Why this
>happens? Is this intended behavior?
Yes, this is the way Partition is supposed to work.
By doing
Partition[list,n,n,-1,x]
you are saying:
pad only at the beginning
pad using n-1 x with the first element of list in the nth
position of the first group
take the next set of n elements for the next group
continue as long as you can take n more elements
stop once you cannot get n elements
By doing
Partition[list,n,n,1,x]
you pad at the end with enough x's to make the flattened length
a multiple of n.
Perhaps you want to pad at both ends so that nothing gets
dropped. If so, do
Partition[list,n,n,{-1,1},x]
keep in mind Partition[list,n,n,k] is simply shorthand for Partition[list,n,n,{k,k}]