How to converting complex numbers to absolute values?

2022 5 29
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更新时间：2022 5 30
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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

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Stephen23
Stephen23 2022-5-30
"This is how data look like."
In reality it looks like this, including alternating lines with empty fields:
Whoever invented that file format is intentionally trying to make everyone else's lives harder. They are not a happy person.
Masoud Taleb
Masoud Taleb 2022-5-30
编辑：Masoud Taleb 2022-5-30
Yes, it is indeed. It is created by a software for further analysis!

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Stephen23
Stephen23 2022-5-30
编辑：Stephen23 2022-5-30
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Ran in:
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]}

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Masoud Taleb
Masoud Taleb 2022-5-30
Thnak you very much @Stephen23
This exactly does the work very fast. I really enjoyed how well written it is :)

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Chunru
Chunru 2022-5-30
编辑：Chunru 2022-5-30
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Ran in:
a = readcell('sample data.csv');
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

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