Reading plot names from folder using gnuplot

Question

I want to make variable "list" containing filenames, variable "i" containing number of datafiles and plot it into single graph. Example of filenames: ConvAut.dat, ConvMoveAut.dat, CutAut.dat ... Here is my try:

list = system("echo $(dir *.dat)")
plot for [1:i] i using 1:((2*i)+(column(2))) w steps tit i

I just found this piece of code but i dont know how to edit it. I have version 4.6 patchlevel 0. Can you help me? Thanks for the answers

Christoph · Accepted Answer · 2015-09-02 17:33:10Z

1

list = system("dir /b *.dat") # Windows
# list = system("ls *.dat") # Unix
plot for [i=1:words(list)] word(list, i) using 1:((2*i)+column(2)) w steps title word(list, i)

The list is a string which contains all files, separated by white spaces. In the plot command you could also iterate over this file list with plot for [file in list] file ..., but then you don't have access to the file number. So I used words to get the number of files, and word(list, i) to get the i-th file name.

Note, that this works only if the file names don't contain white spaces.

answered Sep 2, 2015 at 17:33

Christoph

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2 Comments

vagoberto Over a year ago

In Unix, one way to handle file names containing white spaces is list = system("ls *.txt | xargs -I line echo \\\"line\\\" ")

Christoph Over a year ago

@vagoberto Thanks for the addition. That would work with gnuplot 5, which has limited support for handling quoted strings with word.

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