# How to converting complex numbers to absolute values?

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Masoud Taleb on 29 May 2022
Edited: Masoud Taleb on 30 May 2022
Hello Matlab experts
I have difficulty with my complex data file. I would appreciate if some one can help me with it. The issue is how to import csv files containing complex values (as seen below) into matlab and then convert them into absolute values and also remove parantesis. I treid several commands but did not succeed it :(
This is how data look like. I also attach it as csv file:
theta\phi[rad] 0j (0.010489457941869092+0j) (0.020978915883738184+0j)(0.00393683289923533+0j) (-1.1834597816970487+0j) (-1.1825728101249893+0j) (-1.1815870828154462+0j)(0.00787366579847066+0j) (-2.498909050828868+0j) (-2.5891527984935+0j) (-2.5591423614638047+0j)(0.01181049869770599+0j) (-2.7942230095552336+0j) (-3.8767443114429523+0j) (-3.7833384009620286+0j)
Thank you very much
##### 2 CommentsShow 1 older commentHide 1 older comment
Masoud Taleb on 30 May 2022
Edited: Masoud Taleb on 30 May 2022
Yes, it is indeed. It is created by a software for further analysis!

Stephen23 on 30 May 2022
Edited: Stephen23 on 30 May 2022
No need for slow fiddling around with imported text via CELLFUN or evil EVAL.
Just specify the delimiter&whitespace and MATLAB will correctly import those complex numbers as numeric. Note that I modified the file (attached) by making a few of the imaginary numbers non-zero.
M = readmatrix('sample data.csv', 'Delimiter',',', 'Whitespace','()')
M =
NaN + 0.0000i 0.0000 + 0.0000i 0.0105 + 0.0000i 0.0210 + 0.0000i 0.0315 + 0.0000i 0.0420 + 0.0000i 0.0524 + 0.0000i 0.0000 + 1.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0039 + 0.0000i -1.1835 + 2.0000i -1.1826 + 0.0000i -1.1816 + 0.0000i -1.1805 + 0.0000i -1.1793 + 0.0000i -1.1780 + 0.0000i 0.0079 + 0.0000i -2.4989 + 0.0000i -2.5892 + 3.0000i -2.5591 + 0.0000i -2.5653 + 0.0000i -2.5740 + 0.0000i -2.5823 + 0.0000i 0.0118 + 0.0000i -2.7942 + 0.0000i -3.8767 + 0.0000i -3.7833 + 4.0000i -3.7168 + 0.0000i -3.6653 + 0.0000i -3.5654 + 0.0000i 0.0157 + 0.0000i -2.5550 + 0.0000i -4.6421 + 0.0000i -5.1153 + 0.0000i -5.1818 + 5.0000i -4.6118 + 0.0000i -4.6674 + 0.0000i
Then you can easily get the absolute value:
A = abs(M)
A = 6×7
NaN 0 0.0105 0.0210 0.0315 0.0420 0.0524 1.0000 0 0 0 0 0 0 0.0039 2.3239 1.1826 1.1816 1.1805 1.1793 1.1780 0.0079 2.4989 3.9628 2.5591 2.5653 2.5740 2.5823 0.0118 2.7942 3.8767 5.5058 3.7168 3.6653 3.5654 0.0157 2.5550 4.6421 5.1153 7.2008 4.6118 4.6674
Note that you could use READCELL, if that first element is important (but storing numeric data in a cell array like this makes processing the numeric data more difficult):
C = readcell('sample data.csv', 'Delimiter',',', 'Whitespace','()')
C = 6×7 cell array
{'theta\phi[rad]' } {[ 0]} {[ 0.0105]} {[ 0.0210]} {[ 0.0315]} {[ 0.0420]} {[ 0.0524]} {[0.0000 + 1.0000i]} {[ 0]} {[ 0]} {[ 0]} {[ 0]} {[ 0]} {[ 0]} {[ 0.0039]} {[-1.1835 + 2.0000i]} {[ -1.1826]} {[ -1.1816]} {[ -1.1805]} {[-1.1793]} {[-1.1780]} {[ 0.0079]} {[ -2.4989]} {[-2.5892 + 3.0000i]} {[ -2.5591]} {[ -2.5653]} {[-2.5740]} {[-2.5823]} {[ 0.0118]} {[ -2.7942]} {[ -3.8767]} {[-3.7833 + 4.0000i]} {[ -3.7168]} {[-3.6653]} {[-3.5654]} {[ 0.0157]} {[ -2.5550]} {[ -4.6421]} {[ -5.1153]} {[-5.1818 + 5.0000i]} {[-4.6118]} {[-4.6674]}
Masoud Taleb on 30 May 2022
Thnak you very much @Stephen23
This exactly does the work very fast. I really enjoyed how well written it is :)

Chunru on 30 May 2022
Edited: Chunru on 30 May 2022
a{1} = 'NaN'
a = 6×7 cell array
{'NaN' } {[ 0]} {'(0.010489457941869092+0j)'} {'(0.020978915883738184+0j)'} {'(0.03146837382560728+0j)'} {'(0.04195783176747637+0j)'} {'(0.05244728970934546+0j)'} {[ 0]} {[ 0]} {[ 0]} {[ 0]} {[ 0]} {[ 0]} {[ 0]} {'(0.00393683289923533+0j)'} {'(-1.1834597816970487+0j)'} {'(-1.1825728101249893+0j)' } {'(-1.1815870828154462+0j)' } {'(-1.1805027837737379+0j)'} {'(-1.1793201155129986+0j)'} {'(-1.1780392991030344+0j)'} {'(0.00787366579847066+0j)'} {'(-2.498909050828868+0j)' } {'(-2.5891527984935+0j)' } {'(-2.5591423614638047+0j)' } {'(-2.5652675338416144+0j)'} {'(-2.573976914138426+0j)' } {'(-2.5822604738413624+0j)'} {'(0.01181049869770599+0j)'} {'(-2.7942230095552336+0j)'} {'(-3.8767443114429523+0j)' } {'(-3.7833384009620286+0j)' } {'(-3.7167845156029586+0j)'} {'(-3.6652520844412733+0j)'} {'(-3.5654248689348518+0j)'} {'(0.01574733159694132+0j)'} {'(-2.554964246927875+0j)' } {'(-4.642096167506649+0j)' } {'(-5.1152675003827115+0j)' } {'(-5.1818108617674445+0j)'} {'(-4.611800660922098+0j)' } {'(-4.6674370489549935+0j)'}
b = zeros(size(a));
for i=1:numel(a)
if isnumeric(a{i})
b(i) = a{i};
else
%a{i}
b(i) = eval(a{i});
end
end
b
b = 6×7
NaN 0 0.0105 0.0210 0.0315 0.0420 0.0524 0 0 0 0 0 0 0 0.0039 -1.1835 -1.1826 -1.1816 -1.1805 -1.1793 -1.1780 0.0079 -2.4989 -2.5892 -2.5591 -2.5653 -2.5740 -2.5823 0.0118 -2.7942 -3.8767 -3.7833 -3.7168 -3.6653 -3.5654 0.0157 -2.5550 -4.6421 -5.1153 -5.1818 -4.6118 -4.6674