Skip to main content

import - How to read CDF (Common Data Format) file?


I don't understand, what Import does with CDF.


If I enter



In[17]:= data = Import[ExampleFile]
Out[17]= {"Pose"}

what I can do with returned object? How can I return Pose entry only?


If I try to specify this as element name as string, I am failing:


In[18]:= data = Import[ExampleFile, "Pose"]

During evaluation of In[18]:= Import::noelem: The Import element "Pose" is not present when importing as NASACDF.

Out[18]= $Failed


UPDATE


I noticed, that it showed entity name in quotes. In notebook it looks different:


enter image description here


i.e. w/o quotes.


UDPATE 2


Shared example: https://drive.google.com/open?id=0By8pZ9a2478YOGlhZUlRU3VNb28


UPDATE 3


Apparently Mathematica has a bug here, because the data returned is reshaped incorrectly.


Here is Matlab transcript, reading the same file:



>> data = cdfread(ExampleFile,'Variable', {'Pose'});
>> data = data{1};
>> size(data)

ans =

992 96

>> data(1,1:5)


ans =

1×5 single row vector

-287.2470 64.5002 933.7140 -417.2063 37.0789

>> data(1:5,1)

ans =


5×1 single column vector

-287.2470
-287.6960
-288.1260
-288.5740
-288.9960

An here is Mathematicas:


In[47]:= data = Import[ExampleFile, "Elements"]

data = Import[ExampleFile, "Data"];
Dimensions[data]
data[[1, 1, 1, 1]]

Out[47]= {"Annotations", "Data", "DataEncoding", "DataFormat", \
"Datasets", "Metadata"}

Out[49]= {1, 1, 992, 96}

Out[50]= -287.247


data[[1, 1, 1, Range[5]]]

Out[54]= {-287.247, -287.696, -288.126, -288.574, -288.996}

In[55]:= data[[1, 1, Range[5], 1]]

Out[55]= {-287.247, -497.341, -378.209, -376.513, -370.388}

I.e. although Mathematica indicates the similar shape of data, i.t. fills that shape in different way.



I.e. what Matlab cosiders as row, Mathematica considers as column, and what Mathematica considers as column, is unknown.


So, I still need instructions if I am incorrectly operating with Mathematica's ND arrays, or, if it is Mathematica bug, I need to know, how to recover from it.


UPDATE 4


For now I see the following way to recover original data from corrupted one:


dims = Dimensions[data]
data = ArrayReshape[
data, {dims[[1]], dims[[2]], dims[[4]], dims[[3]]}];
data = Transpose[data, {1, 2, 4, 3}];

Answer



In WL this format is known as NASACDF



In doubt you can always import "Elements" and try each one:


Import["Posing.cdf", "Elements"]


{"Annotations", "Data", "DataEncoding", "DataFormat", "Datasets", "Metadata"}

Import["Posing.cdf", "Data"]

enter image description here


Initially returned {"Pose"} was a list of datasets stored in a file. You can extract specific dataset via:



 Import["Posing.cdf", {"Datasets", "Pose"}]

All that and more can of course be learned from linked documentation page.


Comments

Popular posts from this blog

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...

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...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...