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Re: Export a vector
*To*: mathgroup at smc.vnet.net
*Subject*: [mg108829] Re: Export a vector
*From*: Bill Rowe <readnews at sbcglobal.net>
*Date*: Fri, 2 Apr 2010 05:20:42 -0500 (EST)
On 4/1/10 at 6:00 AM, sagittarius5962 at gmail.com (nt) wrote:
>I have a question on how to export an array from a loop with
>several iterating varibles: for example:
>equation= some equation!
>t0=1;ds1=1
>Do[{t[i]=i*t0};{eq[i]=equation/.t->t[i]};{s1[j]=j*ds1};bet[i,j]=
>r/.FindRoot[eq[i]==0,{s,s1[j]}];betn[i]=Min[bet[i,j]],{i,1,5},{j,1,5}]
>I can give the loop a name like L=Do[....] and write
>Export["file.xls",L] but what I get is the matrix with dimensions of
>i*j.
If you actually tried to Export as you described above and got
what you stated it would be very surprising. The return value
from Do is Null unless a specific Return is used per the
documentation. Setting L = Do[...] and getting what you report
would be a bug.
>can I only export bet[i] which is a vector within the loop to a
>file?
The first thing you need to realize is bet[i] is not a vector
within the loop. None of your code creates a vector or a matrix
(more properly in Mathematica a list).
The notation bet[i] is the function evaluated with the value i.
And since your code does not define values for bet with single
arguments, bet[i] will return unevaluated.
For example, I can define values for f[i,j] as follows:
In[16]:= Table[f[i, j] = RandomInteger[100], {i, 3}, {j, 3}]
Out[16]= {{35, 5, 19}, {22, 87, 75}, {98, 27, 46}}
and to demonstrate f has been assigned values for specific i,j
In[17]:= f[1, 1]
Out[17]= 35
but note,
In[18]:= f[1]
Out[18]= f[1]
since f has not be defined for a single argument. For something
that behaves like a matrix, I would do:
In[19]:= g = Table[RandomInteger[100], {3}, {3}]
Out[19]= {{70, 1, 89}, {17, 92, 49}, {43, 24, 72}}
Now I can extract the first row by:
In[20]:= g[[1]]
Out[20]= {70,1,89}
or the first column by:
In[21]:= g[[All, 1]]
Out[21]= {70,17,43}
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