Incorrect use of '=' operator. To assign a value to a variable, use '='. To compare values for equality, use '=='.

2 visualizzazioni (ultimi 30 giorni)
Hello everyone,
I'm getting the aformentioned error for the RL2 variable but I do not want to compare values for equality. I just want to put RL2 = RL/2. This problem persists even when I remove that line of code. The same problem occurs for H matrix when I remove the line of code for RL2 and the same occurs for R when I remove the line of code for H matrix. Why is this happening?
Thanks
clear all
close all
clc
R1 = 0; R2 = 100; R3 = 125; R4 = 150; R5 = 200; R6 = 250; R7 = 500; R8 = 1000;
R9 = 2000;
L1 = 0; L2 = 2.5; L3 = 5; L4 = 7.5; L5 = 10; L6 = 12.5; L7 = 15; L8 = 17.5; L9 = 20;
RL = [R1*L1 R2*L1 R3*L1 R4*L1 R5*L1 R6*L1 R7*L1 R8*L1 R9*L1;
R1*L2 R2*L2 R3*L2 R4*L2 R5*L2 R6*L2 R7*L2 R8*L2 R9*L2;
R1*L3 R2*L3 R3*L3 R4*L3 R5*L3 R6*L3 R7*L3 R8*L3 R9*L3;
R1*L4 R2*L4 R3*L4 R4*L4 R5*L4 R6*L4 R7*L4 R8*L4 R9*L4;
R1*L5 R2*L5 R3*L5 R4*L5 R5*L5 R6*L5 R7*L5 R8*L5 R9*L5;
R1*L6 R2*L6 R3*L6 R4*L6 R5*L6 R6*L6 R7*L6 R8*L6 R9*L6;
R1*L7 R2*L7 R3*L7 R4*L7 R5*L7 R6*L7 R7*L7 R8*L7 R9*L7;
R1*L8 R2*L8 R3*L8 R4*L8 R5*L8 R6*L8 R7*L8 R8*L8 R9*L8;
R1*L9 R2*L9 R3*L9 R4*L9 R5*L9 R6*L9 R7*L9 R8*L9 R9*L9;
RL2 = (RL/2);
H = [73.00 73.00 73.00 73.00 73.00 73.00 73.00 73.00 73.00;
70.17 70.17 70.17 70.17 70.17 70.17 70.17 70.17 70.17;
67.34 67.34 67.34 67.34 67.34 67.34 67.34 67.34 67.34;
64.19 64.19 64.19 64.19 64.19 64.19 64.19 64.19 64.19;
61.36 61.36 61.36 61.36 61.36 61.36 61.36 61.36 61.36;
58.53 58.53 58.53 58.53 58.53 58.53 58.53 58.53 58.53;
55.38 55.38 55.38 55.38 55.38 55.38 55.38 55.38 55.38;
49.72 49.72 49.72 49.72 49.72 49.72 49.72 49.72 49.72;
43.74 43.74 43.74 43.74 43.74 43.74 43.74 43.74 43.74;
37.76 37.76 37.76 37.76 37.76 37.76 37.76 37.76 37.76;
32.09 32.09 32.09 32.09 32.09 32.09 32.09 32.09 32.09;
23.60 23.60 23.60 23.60 23.60 23.60 23.60 23.60 23.60;
15.73 15.73 15.73 15.73 15.73 15.73 15.73 15.73 15.73;
7.87 7.87 7.87 7.87 7.87 7.87 7.87 7.87 7.87;
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00];
R = [R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9;
R1 R2 R3 R4 R5 R6 R7 R8 R9];
contourf(RL2,R,H)
k = colorbar;
cmap = colormap;
ax = gca;
ax.Color = cmap(1, :);
k.Label.String = 'Relative Humidity (%)';
xlabel('Length (\mum)')
ylabel('Radius (\mum)')
  2 Commenti
Ruslan Smirnov
Ruslan Smirnov il 5 Ott 2022
Modificato: Ruslan Smirnov il 5 Ott 2022
surprisingly, when trying to execute code section by section, problems with incomplete lines or wrong usage of = operator affect the execution of previous lines; pathetic
Steven Lord
Steven Lord il 5 Ott 2022
@asd ad Rather than defining individual variables named R1 through R9 and L1 through L9, I recommend defining a vector.
R = 1:9;
L = 10:19;
With vectors and implicit expansion, you can assemble those matrices with a lot less typing.
A = R.*(L.')
A = 10×9
10 20 30 40 50 60 70 80 90 11 22 33 44 55 66 77 88 99 12 24 36 48 60 72 84 96 108 13 26 39 52 65 78 91 104 117 14 28 42 56 70 84 98 112 126 15 30 45 60 75 90 105 120 135 16 32 48 64 80 96 112 128 144 17 34 51 68 85 102 119 136 153 18 36 54 72 90 108 126 144 162 19 38 57 76 95 114 133 152 171
As a check, A(3,4) should be R(4)*L(3).
[A(3,4) R(4)*L(3)]
ans = 1×2
48 48

Accedi per commentare.

Risposta accettata

Walter Roberson
Walter Roberson il 14 Ago 2020
RL2 = (RL/2;
That is an incomplete expression that would cause problems for what follows.
  4 Commenti
Walter Roberson
Walter Roberson il 14 Ago 2020
After you put in the ] for RL then there are no more syntax errors, but you have the problem that in your call
contourf(RL2,R,H)
then RL2 is 9 x 9 but R and H are 15 x 9

Accedi per commentare.

Più risposte (0)

Tag

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by