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}