Problem with if statements

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

Answer 1

[David Hill]

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

[Blake]

Thanks so much. Could you explaing whan the ismember, sting, and abs do? Thanks again.

[David Hill]

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

Answer 2

[Steven Lord]

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]

This is amazing and extremely helpful, thank you so much.