# Why does my <= statement not work?

6 views (last 30 days)
HC98 on 31 Jan 2020
Edited: Stephen23 on 31 Jan 2020
I'm running a simple loop that applies a less than or equal to inequality to a value. Unfortunately, it is overlooking this value and not working. Here is my code
% Initial Conditions
clearvars
close all
Ms=1.989E30;
m = logspace(-2, 2, 400);
% m = 1E2:100:1E4
% (m<0.08, m**-0.3, numpy.where(m < 0.5, 0.08**-0.3 * (m/0.08)**-1.3, 0.08**-0.3 * (0.5/0.08)**-1.3 * (m/0.5)**-2.3))
%% def chabrier03individual(m):
k = 0.158*exp(-(-log(0.08))^2/(2*0.69^2))
if m<=1
u = 0.158*(1./m)*exp(-(log(m)-log(0.08))^2/(2 * 0.69^2));
else
v = k*m;
end
Stephen23 on 31 Jan 2020
Edited: Stephen23 on 31 Jan 2020
The comprison works as documented. Note the if documentation states " An expression is true when its result is nonempty and contains only nonzero elements (logical or real numeric). Otherwise, the expression is false."
You provided if with a 1x400 logical vector as its condition, it contains no non-zero elements:
>> nnz(m<=1)
ans = 0
Based on this I would expect the else part to be executed, and this is indeed what occurs when i run your code.
Note that your vector m contains just one value repeated 400 times over:
>> unique(m)
ans = 100
It is not clear why you wanted to create that 1x400 vector in such an obfuscated way.
EDIT: the OP originally defined
m = logspace(2, 2, 400);
but has since edited the question and changed it to:
m = logspace(-2, 2, 400);
The relevance of reading the if documentation is unchanged.

Shekhar Vats on 31 Jan 2020
Stephen23 on 31 Jan 2020
Edited: Stephen23 on 31 Jan 2020
The equivalent to numpy's where is to use indexing:
x = ...;
y = ...;
idx = m<=1;
z = y;
z(idx) = x(idx)