While loop - display the last value while statement is still true

The example given on Mathworks doesn't make sense to me.
x=0
p=1-poisscdf(1,x);
while p<=0.1
x=x+0.01;
p=1-poisscdf(1,x);
end
display(x);
The x value given at the end of the loop is the first value that makes the statement false.What can I add to the script so that display(x) gives me the last value that makes the statement true?
*Edit:
I just tried:
x=0;
p=1-poisscdf(1,x);
while p2more<=0.1
x=x+0.01;
p=1-poisscdf(1,x);
if p>=0.1
break
end
display(x);
end
Now I get all the values of x where this statement is true. I just want the last value. :X

 Risposta accettata

You could just use another variable. E.g., y
x=0
p=1-poisscdf(1,x);
while p<=0.1
y = x; % <-- save current value of x
x=x+0.01;
p=1-poisscdf(1,x);
end
display(y); % <-- display last value of x for which test was true

3 Commenti

Thank you! I accepted this since it was posted earlier, but Jan Simon also provided this solution. Thumbs up to you both.
If it's not too much trouble could you explain why using another variable got me to the last value of x for which the test was true?
Right after the test passes, you save the x value that made the test pass. But when you generate an x value that doesn't pass the test, you don't execute the y = x statement. So it is only the x values that pass the test that are saved in y. And the last one is what makes it to the display(y) statement.
Thank you, that was really helpful.

Accedi per commentare.

Più risposte (1)

Jan
Jan il 21 Mar 2017
Modificato: Jan il 21 Mar 2017
The trivial soultion is:
x = 0;
p = 1 - poisscdf(1, x);
while p <= 0.1
x = x + 0.01;
p = 1 - poisscdf(1, x);
end
x = x - 0.01;
display(x);
If the increasing of x is not a constant:
x = 0;
lastx = x;
p = 1 - poisscdf(1, x);
while p <= 0.1
lastx = x;
x = x + 0.01;
p = 1 - poisscdf(1, x);
end
display(lastx);
Now p is defined twice and redundacies in the code are prone to bugs: If one instance is modified later, that other might be forgotton. To avoid this:
x = 0;
ready = false
while ~ready
p = 1 - poisscdf(1, x);
ready = (p < 0.1);
lastx = x;
x = x + 0.01;
end
display(lastx);

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Richiesto:

il 20 Mar 2017

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il 24 Mar 2017

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