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I wanna give names to transitions in my file, but whenever I want to name something it changes to a numerical value or NaN.
So whenever the value is -1 -> P, 0 -> Q and 1 -> R.
FileName = 'ct4004355_si_003.txt';
T = readtable('ct4004355_si_003.txt', 'Delimiter','\t', 'TextType','string');
Rows = T(130:154,[1 2 3 4 5 6 11 12 13 14]);
Rows.Properties.VariableNames = ["Wavenumber [cm-1]", "Uncertainty", "v1u", "v2u", "L2u", "Ju", "v1l", "v2l", "L2l", "Jl"];
PR = Rows.Ju-Rows.Jl;
Rows = [Rows table(PR, 'VariableNames', {'Branch'}) ];
Rows.Branch(ismember(Rows.Branch, [-1])) = 'P'

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Ben van Zon
Ben van Zon 2023-9-14
25×11 table
Wavenumber [cm-1] Uncertainty v1u v2u L2u Ju v1l v2l L2l Jl Branch
_________________ ___________ ___ ___ ___ __ ___ ___ ___ __ ______
2051.5 0.01 0 1 1 5 0 0 0 6 80
2054 0.01 0 2 2 5 0 1 1 6 80
2057.4 0.01 0 2 0 1 0 1 1 2 80
2060.2 0.01 0 1 1 6 0 0 0 7 80
2061.7 0.01 0 1 1 5 0 0 0 6 80
: : : : : : : : : : :
2134.2 0.01 0 2 0 7 0 1 1 7 0
2134.9 0.01 0 1 1 4 0 0 0 5 80
2137 0.01 0 2 2 4 0 1 1 5 80
2140.3 0.01 0 1 1 4 0 0 0 5 80
2142.3 0.01 0 2 2 4 0 1 1 5 80
This is the result, instead of '80' I need a 'P'.

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 采纳的回答

Dyuman Joshi
Dyuman Joshi 2023-9-14
Ran in:

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FileName = 'ct4004355_si_003.txt';
T = readtable('ct4004355_si_003.txt', 'Delimiter','\t', 'TextType','string');
Rows = T(130:154,[1 2 3 4 5 6 11 12 13 14]);
Rows.Properties.VariableNames = ["Wavenumber [cm-1]", "Uncertainty", "v1u", "v2u", "L2u", "Ju", "v1l", "v2l", "L2l", "Jl"];
PR = Rows.Ju-Rows.Jl;
Rows = [Rows table(PR, 'VariableNames', {'Branch'}) ];
str = 'PQR';
Rows.Branch = str(Rows.Branch+2)'
Rows = 25×11 table
Wavenumber [cm-1] Uncertainty v1u v2u L2u Ju v1l v2l L2l Jl Branch _________________ ___________ ___ ___ ___ __ ___ ___ ___ __ ______ 2051.5 0.01 0 1 1 5 0 0 0 6 P 2054 0.01 0 2 2 5 0 1 1 6 P 2057.4 0.01 0 2 0 1 0 1 1 2 P 2060.2 0.01 0 1 1 6 0 0 0 7 P 2061.7 0.01 0 1 1 5 0 0 0 6 P 2067.4 0.01 0 1 1 6 0 0 0 7 P 2074 0.04 0 2 2 5 0 1 1 6 P 2077.5 0.01 0 1 1 6 0 0 0 7 P 2079.4 0.01 0 1 1 5 0 0 0 6 P 2080.7 0.03 0 1 1 6 0 0 0 7 P 2089.3 0.01 1 1 1 3 1 0 0 4 P 2089.8 0.01 0 2 2 5 0 1 1 6 P 2094.2 0.015 0 2 2 5 0 1 1 6 P 2095.3 0.02 0 2 2 4 0 1 1 5 P 2096.6 0.01 0 1 1 5 0 0 0 6 P 2097.7 0.01 1 1 1 3 1 0 0 4 P

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Ben van Zon
Ben van Zon 2023-9-14
First of all, thank you!
Second, how does that work so simply?
Dyuman Joshi
Dyuman Joshi 2023-9-14
Ran in:
Ran in:
> Explaination - When you convert the data type of a portion of data, (if the conversion is allowed) MATLAB retains the data type and changes the modified data to the original data type.
%Double numeric
y = 1:5
y = 1×5
1 2 3 4 5
z = 'Yo'
z = 'Yo'
%Last 2 values changed to string, but the MATLAB converts the corresponding
%string to double numeric
y(end-1:end) = z
y = 1×5
1 2 3 89 111
%Character vector
y = 'Maths is fun'
y = 'Maths is fun'
%1st 3 values changed to numeric, MATLAB converts the
%corresponding numbers to their character (ASCII) values
y(1:3) = [69 420 666]
y = 'EƤʚhs is fun'
If you want to change the data-type of variable corresponding to some value, over-write the whole array -
y = 1:5;
y = sprintf('%d', y)
y = '12345'
In your case, the original data was numeric. You modified some values to be a character ('P' to be exact), MATLAB converted it to the numerical value
double('P')
ans = 80
I over-wrote the whole data.
> P = -1, Q = 0, R = 1
can be changed to
P+2 = 1, Q+2 = 2, R+2 = 3
So, I used 1, 2, and 3 as indices, corresponding to 'P', 'Q' and 'R'.
Why change? Because indexing in MATLAB starts with 1.
> What if the values were different and the data was much bigger? Say -
P = -53, Q = 13, R = 0
%vec is only used to generate data
vec = [-53 0 13];
str = 'PQR';
%And this is the data we are working with
data = vec(randi(3,1,100))
data = 1×100
0 0 0 0 -53 -53 -53 -53 13 13 -53 13 0 0 -53 0 -53 0 13 0 0 13 -53 0 13 -53 0 13 13 0
%Then get the unique values and corresponding indices
[arr,~,idx] = unique(data)
arr = 1×3
-53 0 13
idx = 100×1
2 2 2 2 1 1 1 1 3 3
%And use the indices to change the values
data = str(idx)
data = 'QQQQPPPPRRPRQQPQPQRQQRPQRPQRRQPRRQQQQPQPPPRRPRRPRRRQQRQPRQQPRRQRPRPRQPRQQRQPQQPQRRRQRQPPQRQQPQRRQPQR'
Ben van Zon
Ben van Zon 2023-9-14
Thank you for the extensive answer!
This will be useful for the other things I'm trying to do with this table.

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