# Problem with if statements

1 view (last 30 days)
Blake on 21 Nov 2022
Commented: David Hill on 28 Nov 2022
Hey!
Whenever I run this script, the answer polar is always displayed even though the answer should be nonpolar. Any help would be apprecitated. Thank you.
H = 2.20;
Li = 0.98;
Be = 1.57;
B = 2.04;
C = 2.55;
N = 3.04;
O = 3.44;
F = 3.98;
Na = 0.93;
Mg = 1.31;
Al = 1.61;
Si = 1.90;
P = 2.19;
S = 2.58;
Cl = 3.16;
K = 0.82;
Ca = 1;
Se = 1.36;
Ti = 1.54;
V = 1.63;
Cr = 1.66;
Mn = 1.55;
Fe = 1.83;
Co = 1.88;
Ni = 1.91;
Cu = 1.90;
Zn = 1.65;
Ga = 1.81;
Ge = 2.01;
As = 2.18;
Se = 2.55;
Br = 2.96;
FirstElement = input('What is the first element in the compound?');
SecondElement = input('What is the second element in the compound?');
Polarity = SecondElement - FirstElement;
while Polarity == SecondElement - FirstElement;
if FirstElement == Li, SecondElement == H;
Polarity = H - Li;
end
if FirstElement == Be, SecondElement == H;
Polarity = H - Be;
end
if FirstElement == B, SecondElement == H;
Polarity = H - B;
end
if FirstElement == B, SecondElement == H;
Polarity = H - B;
end
if FirstElement == B, SecondElement == C;
Polarity = C - B;
end
if FirstElement == B, SecondElement == N;
Polarity = N - B;
end
if FirstElement == B, SecondElement == O;
Polarity = O - B;
end
if FirstElement == B, SecondElement == F;
Polarity = F - B;
end
end
disp(ans(Polarity))
if Polarity > 0.4;
disp('Polar')
elseif disp('Non Polar')
end

David Hill on 21 Nov 2022
e=["H","Li","Be","B","C","N","O","F","Na","Mg","Al","Si","P","S","Cl","K","Ca",...
"Sc","Ti","V","Cr","Mn","Fe","Co","Ni","Cu","Zn","Ga","Ge","As","Se","Br"];
l=[2.20;0.98;1.57;2.04;2.55;3.04;3.44;3.98;0.93;1.31;1.61;1.90;2.19;2.58;3.16;...
0.82;1;1.36;1.54;1.63;1.66;1.55;1.83;1.88;1.91;1.90;1.65;1.81;2.01;2.18;...
2.55;2.96];
FirstElement = find(ismember(e,string(input('What is the first element in the compound?','s'))));
SecondElement = find(ismember(e,string(input('What is the second element in the compound?','s'))));
Polarity = abs(l(SecondElement) - l(FirstElement));
if Polarity > 0.4
disp('Polar')
else
disp('Non Polar')
end
##### 2 CommentsShowHide 1 older comment
David Hill on 28 Nov 2022
ismember is producing a logical array the same size as e. You could also do e=="Cr" which produces the same logical array. abs is just the absolute value of the subtraction. input will produce a character array but we need to compare string to string. string just converts the character array into a string.
e=["H","Li","Be","B","C","N","O","F","Na","Mg","Al","Si","P","S","Cl","K","Ca",...
"Sc","Ti","V","Cr","Mn","Fe","Co","Ni","Cu","Zn","Ga","Ge","As","Se","Br"];
ismember(e,"Cr")
ans = 1×32 logical array
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
e=="Cr"
ans = 1×32 logical array
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0

Steven Lord on 21 Nov 2022
Edited: Steven Lord on 21 Nov 2022
Don't define a large number of variables whose names match the symbols for the elements and use a large collection of if statements. Instead I recommend using either a struct array with dynamic field names or a table with named rows.
polarityValues = struct(H = 2.20, Li = 0.98, Be = 1.57, B = 2.04)
polarityValues = struct with fields:
H: 2.2000 Li: 0.9800 Be: 1.5700 B: 2.0400
Or if you have too many fields to set this way, you could define them with lines like:
polarityValues.C = 2.55
polarityValues = struct with fields:
H: 2.2000 Li: 0.9800 Be: 1.5700 B: 2.0400 C: 2.5500
Now if we want to retrieve the polarityValues for two elements:
firstElement = 'Li';
secondElement = 'Be';
polarity(1) = polarityValues.(firstElement);
polarity(2) = polarityValues.(secondElement)
polarity = 1×2
0.9800 1.5700
For a table-based approach:
elementSymbols = ["H"; "Li"; "Be"; "B"; "C"];
polarityValues = [2.2; 0.98; 1.57; 2.04; 2.55];
t = table(polarityValues, 'VariableNames', "Polarity", 'RowNames', elementSymbols)
t = 5×1 table
Polarity ________ H 2.2 Li 0.98 Be 1.57 B 2.04 C 2.55
To retrieve values:
polarity = t{'Li', 'Polarity'}
polarity = 0.9800
polarity = t{{firstElement, secondElement}, 'Polarity'}
polarity = 2×1
0.9800 1.5700
This neatly avoids the problem where you don't have an if statement for a particular combination of elements, in which case the polarity variable will be defined as the difference between the symbols of the elements due to your line "Polarity = SecondElement - FirstElement;"
polarity = firstElement - secondElement
polarity = 1×2
10 4
L is 10 characters later in the alphabet than B and i is 4 characters later in the alphabet than e. Because in this case polarity is a vector, the body of that last if statement in your code will be executed if all the elements of the vector are greater than 0.4, and in this case they are.
Blake on 24 Nov 2022
This is amazing and extremely helpful, thank you so much.

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