Skip to main content

list manipulation - Selecting multiple columns from a matrix?


Sample data:


data = {
{{2013, 1, 1}, 24.13, 167.67, 231.82},
{{2013, 1, 2}, 32.15, 170.92, 225.99},

{{2013, 1, 3}, 35.43, 172.68, 221.67},
{{2013, 1, 4}, 36.73, 173.05, 218.32},
{{2013, 1, 5}, 58.19, 165.96, 197.05},
{{2013, 1, 6}, 69.99, 163.50, 187.52},
{{2013, 1, 7}, 71.37, 154.21, 175.58},
{{2013, 1, 8}, 72.51, 149.66, 163.25}};

I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code:


data2 = Transpose[{data[[All, 1]], data[[All, 2]]}];
data3 = Transpose[{data[[All, 1]], data[[All, 3]]}];

data4 = Transpose[{data[[All, 1]], data[[All, 4]]}];
DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}]

but I have a hunch that this can be done more efficiently. I don't like the Transposes in particular. Any ideas?



edit (for extra credit)
What if I need to multiply the second column by 2, which in my solution is simply


data2 = Transpose[{data[[All, 1]], 2 * data[[All, 2]]}];

Answer



This method allows you to define arbitrary column indexes, not necessarily that orderly as you have. Just put them in the list:



iNeed = {{1, 2}, {1, 3}, {1, 4}};

then this will do the job and make a plot identical to yours:


DateListPlot[data[[All, #]] & /@ iNeed, Joined -> True, Filling -> {3 -> {1}}]

basically it does this but in a shorter way:


DateListPlot[{
data[[All, {1, 2}]],
data[[All, {1, 3}]],
data[[All, {1, 4}]]

}, Joined -> True, Filling -> {3 -> {1}}]

Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]