character encoding - Importing and dealing with foreign languages

I am importing a text file with Swedish letters (å, ä, ö).

A. If I use ReadList["sv_JSPFirefox.txt"], it imports the file nicely but then I cannot use a command line like:

dict2 = Select[dict1, Not@StringContainsQ[#, Alternatives @@ rejectlist] &]

The error code is:

StringContainsQ::strse: String or list of strings expected at position 1 
in StringContainsQ[Aapua,.|0|1|2|3|4|5|6|7|8|9|{A,B,C,D,E,F,G,H,

I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z}|Å|Ä|Ö]. >>

Two screen dumps show the problem better:

B. If I use: ReadList["sv_JSPFirefox.txt, Word] it imports the file but those characters are a mess; but at least I can use the command above and proceed; but commands (search, etc) including these characters do work properly.

I have tried to add "UTF8", "UTF-8","Unicode", "Lines" and other options but nothing seems to help. Any pointer in the right direction would be greatly appreciated. Thank you for your time!

20160306 Edit. @C.E. I made a sample file and a very small file called "shrimpsandwich.txt". This is enough because this is a word in Swedish that contains all three letters, "räksmörgås". You might have heard of smörgåsbord.

Sample files: A small sample file is displayed in the screen dump above.

Answer

As you can see from the discussion above which might give some other insights, the answer by @Xavier solves the problem. Thank you all, for taking the time to test and answer.

Solution:

dict1 = ToString /@ ReadList["sv_JSPFirefox.txt"]

Comments

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}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....

Blog

Search This Blog