I imported the 3rd sheet in a xlsx file, which contains 3 columns of data, using
import["e:/temp/15c.xlsx"][[3]]
The data looks like {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}. I wanted to do curve fitting for just one column of data, such as {1, 4, 7} or {2, 5, 8}. I tried
Fit[%[[1, All, All]], {1, x, x^2}, {x}]
but it didn't work. Would anyone please suggest how to get any 1D data out of the 3D table imported? Thank you.
In[1]:= $list = {{1,2,3},{4,5,6},{7,8,9}};
In[2]:= $list[[All, 1]]
Out[2]= {1, 4, 7}
In[3]:= $list[[All, 2]]
Out[3]= {2, 5, 8}
In[4]:= $list[[All, 3]]
Out[4]= {3, 6, 9}
In[5]:= Fit[$list[[All, 2]], {1, x, x^2}, x] // Chop
Out[5]= -1. + 3. x
Related
In Mathematica - how do I bin an array to create a new array which consist from sum domains of the old array with a given size ???
Example:
thanks.
This is slightly simpler than #ChrisDegnen's solution. Given the same definition of array the expression
Map[Total, Map[Flatten, Partition[array, {2, 2}], {2}], {2}]
produces
{{4, 10}, {8, 10}}
If you prefer, this expression
Apply[Plus, Map[Flatten, Partition[array, {2, 2}], {2}], {2}]
uses Apply and Plus rather than Map and Total but is entirely equivalent.
This works for the example but a generalised version would need more work.
array =
{{1, 1, 1, 2},
{1, 1, 3, 4},
{2, 2, 2, 3},
{2, 2, 2, 3}};
Map[Total,
Map[Flatten,
Map[Transpose,
Map[Partition[#, 2] &, Partition[array, 2], 2],
2], {2}], {2}]
% // MatrixForm
4 10
8 10
I have a list "data1":
{{1, 6, 4.5, 1, 141.793, 2.31634, 27.907}, {2, 7, 4.5, 1, 133.702,
2.28725, 26.7442}, {3, 5, 5, 1, 136.546, 2.33522, 25.5814}, {4, 8,
5, 1, 104.694, 2.27871, 24.4186}}
What I would like to do is to create a new table with only the first two columns of each element. So my new table would be:
{{1,6},{2,7},{3,5},{4,8}}
I tried
data1[[All, 1][All, 2]]
and other variations but I am not understanding how to capture the desired fields. Thank you for your help.
Just have a range or list of the indices you want as the second argument, like so:
In[71]:= data[[All, {1, 2}]]
Out[71]= {{1, 6}, {2, 7}, {3, 5}, {4, 8}}
For example, I have a list like:
{{1, 2, 3}, {6}, {4, 5}, {1, 6}, {2, 2, 3, 2}, {9}, {7}, {2, 5}}
And I want to get a tallied list based on the total of the lists' elements.
In this case, I want the output to be:
{{6, {{1, 2, 3}, {6}}, {7, {{2, 5}, {1, 6}, {7}}}, {9, {{4, 5}, {2, 2, 3, 2}, {9}}}}}
How to do this conveniently in Mathematica?
Thanks a lot.
Here's my attempt - a little simpler than Yoda's
lst = {{1, 2, 3}, {6}, {4, 5}, {1, 6}, {2, 2, 3, 2}, {9}, {7}, {2, 5}};
{Total#First##, #} & /# GatherBy[lst, Total]
If you don't want repeated elements, then you could use
{Total#First##, Union[#]} & /# GatherBy[lst, Total]
Or if you really wanted a tally-like operation
{Total#First##, Tally[#]} & /# GatherBy[lst, Total]
While I would probably do this just as #Simon did, let us not forget that Reap and Sow can be used as well:
Reap[Sow[#, Total[#]] & /# lst, _, List][[2]]
where lst is the original list. This will be somewhat less efficient than the GatherBy- based code, but also pretty fast. One can speed up the above code about 1.5 times by rewriting it as
Reap[Sow ### Transpose[{lst, Total[lst, {2}]}], _, List][[2]]
in which case it becomes about 1.5 times slower than the code based on GatherBy. Note that the speed difference between the two methods is not very dramatic here, because the list is ragged and therefore not packed, and GatherBy does not have here the speed advantage it normally enjoys for packed arrays.
Don't overlook Tr. This is shorter and faster:
{Tr##, {##}} & ### GatherBy[lst, Tr]
Consider :
Tuples[Range[1, 3], 2]
I would like to drop some of the sublist based on the following list :
sublistToTemove = {1,2,3,6,8}
Desired Output :
{2, 1}, {2, 2}, {3, 1}
Corresponding to the 4th, 5th and 7th elements of list.
I have tried Drop, Case, Select without success, must be missing something.
Given your list:
In[2]:= lst = Tuples[Range[1, 3], 2]
Out[2]= {{1, 1}, {1, 2}, {1, 3}, {2, 1}, {2, 2}, {2, 3}, {3, 1}, {3,2}, {3, 3}}
and
In[5]:= sublistToTemove = {1, 2, 3, 6, 8}
Out[5]= {1, 2, 3, 6, 8}
Here are 2 ways:
In[6]:= Delete[lst, List /# sublistToTemove]
Out[6]= {{2, 1}, {2, 2}, {3, 1}, {3, 3}}
In[7]:= lst[[Complement[Range[Length[lst]], sublistToTemove]]]
Out[7]= {{2, 1}, {2, 2}, {3, 1}, {3, 3}}
In[15]:= sublistToTemove = {1, 2, 3, 6, 8};
In[16]:= Delete[Tuples[Range[1, 3], 2], Transpose[{sublistToTemove}]]
Out[16]= {{2, 1}, {2, 2}, {3, 1}, {3, 3}}
I have a set of regular (mod 5) matrices N2 and I would like to get the group generated by these matrices in Mathematica:
My approach was to use a function f doing the matrix multiplication and g for mod 5 and then I wanted to use MapThread
M= Function[{x,y},x.y];
g = Function[z, Mod[z, 5]]
g /# MapThread[M, {N2,N2}]
The problem is that MapThread is inserting only pairs of elements that are at the same position in the lists. I would like to insert any pair of elements in N. To get the group generated by the matrices in N I would just repeat this and update N2 every time.
E.g. let N2 ={A,B}
g /# MapThread[M, {N2,N2}]
would return {B^2,A^2}, while I want it to return any product of matrices in N2, i.e. {A^2,AB,BA,B^2}.
I'm not sure whether I understand your question, but if your intention is to get all combinations of the two matrices A,B you could use Tuples combined with Apply (which you may use in its functional form with square brackets or as many here do in initially cryptic prefix operator form ### = Apply at level 1):
In[24]:= Dot ### Tuples[{A, B}, 2]
Out[24]= {A.A, A.B, B.A, B.B}
In this case you need Outer:
In[27]:= n = RandomInteger[{1, 5}, {3, 2, 2}];
In[28]:= Outer[mFunc, n, n, 1]
Out[28]= {{mFunc[{{3, 5}, {2, 4}}, {{3, 5}, {2, 4}}],
mFunc[{{3, 5}, {2, 4}}, {{3, 4}, {4, 3}}],
mFunc[{{3, 5}, {2, 4}}, {{4, 4}, {5, 1}}]}, {mFunc[{{3, 4}, {4,
3}}, {{3, 5}, {2, 4}}],
mFunc[{{3, 4}, {4, 3}}, {{3, 4}, {4, 3}}],
mFunc[{{3, 4}, {4, 3}}, {{4, 4}, {5, 1}}]}, {mFunc[{{4, 4}, {5,
1}}, {{3, 5}, {2, 4}}],
mFunc[{{4, 4}, {5, 1}}, {{3, 4}, {4, 3}}],
mFunc[{{4, 4}, {5, 1}}, {{4, 4}, {5, 1}}]}}
In[29]:= n
Out[29]= {{{3, 5}, {2, 4}}, {{3, 4}, {4, 3}}, {{4, 4}, {5, 1}}}