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Re:Need help with creating a List
*To*: mathgroup at yoda.ncsa.uiuc.edu (Mathematica Mailing List)
*Subject*: Re:Need help with creating a List
*From*: uunet!kinghorn.chem.WSU.EDU!don (Don Kinghorn)
*Date*: Fri, 24 Aug 90 10:40:23 PDT
Begin recieved message:
************************************************************************************
. . .
I am having some trouble with creating a list for plotting with
ListPlot. As I understand it, what is needed is a list of lists, where
each element of the main list is a list of two values, the x and y
positions to be plotted. In my case, the x values are just a set of
evenly spaced values, while the y values are the difference between some
data I have and a functional form for the data. What I have is:
func[x_] = "some long function"
The data is in a list called "data"
diff[x_] := data[[x]] - func[.05x]
difftable = Table[ {.05i, diff[i]}, {i, Length[data]} ]
My problem is the output of difftable. It looks like:
{{.05, {value1}}, {.01, {value2}}, ....}
What it needs to be for ListPlot to work (I think) is:
{{.05, value1}, {.01, value2}, ....}
I tried to make a test file with a simple func[x_] and just put in some
datapoints, and there it did work. So I'm confused as to when and why
mathematica is giving me {{xval1,yval1},{xval2,yval2}, ..} in one case
and {{xval1,{yval1}}, {xval2,{yval2}}, ..} in the other case.
Any ideas or suggestions?
--Avi Feldblum
ayf at pruxk.att.com or avi_feldblum at att.com
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Look at your y data. I suspect that its in the form of a list of lists ( an nx1 matrix)
and as a result your function diff is returning a vector. Try applying Flatten[]
to your y data to put it in the form {y1,y2,y3, . . .}. This should give the result
you want.
Example:
In[1]:= data=Transpose[{{1,2,3,4}}]
Out[1]= {{1}, {2}, {3}, {4}}
In[2]:= fcn[x_] := x-1/2
In[3]:= dif[x_] := data[[x]]-fcn[.05x]
In[4]:= Table[{.05i,dif[i]},{i,Length[data]}]
Out[4]= {{0.05, {1.45}}, {0.1, {2.4}}, {0.15, {3.35}}, {0.2, {4.3}}}
In[5]:= data = Flatten[data]
Out[5]= {1, 2, 3, 4}
In[6]:= Table[{.05i,dif[i]},{i,Length[data]}]
Out[6]= {{0.05, 1.45}, {0.1, 2.4}, {0.15, 3.35}, {0.2, 4.3}}
I hope this helps. (this is only one of many possible solutions)
Don Kinghorn
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