Write a function called freezing that takes a vector of numbers that correspond to daily low temperatures in Fahrenheit. Return numfreeze, the number of days with sub freezing temperatures (that is, lower than 32 F) without using loops. Here is an ex

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This question is soft-locked: new answers that are equivalent to already posted answers may be deleted without prior notice.
Hello there, I am very new to Matlab and I am having trouble with this question. I understand how to make the function work for the given matrix in the problem. However, I cannot find out how to make it work for random temperature vectors. Would anyone mind giving me a hint or helping me out? Would be greatly appreciated. Thank you.
  3 个评论
Andrew Marttini
Andrew Marttini 2019-7-12
编辑:Andrew Marttini 2019-7-12
So this is what I have so far. Its mostly just the code to solve for the vector thats given.
function numfreeze = freezing (n)
n = [45 21 32 31 51 12]
n1 = n(n<32)
numfreeze = numel(n1)
end

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采纳的回答

Stephan
Stephan 2019-7-12
function numfreeze = freezing (n)
n1 = n(n<32)
numfreeze = numel(n1)
end
Dont overwrite n - it is an input argument
  7 个评论

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更多回答(6 个)

Vineet Singhal
Vineet Singhal 2019-10-14
function numfreeze = freezing(v)
a= length(v(v<32));
numfreeze =a;
end

Nadeem U Rehman
Nadeem U Rehman 2020-12-10
function numfreeze = freezing(A)
F = A(A<32);
[row column] = size(A);
if size(A) == [1 column]
numfreeze = size(F,2);
else
numfreeze = size(F,1);
end
end
  4 个评论
Nadeem U Rehman
Nadeem U Rehman 2020-12-10
编辑:Nadeem U Rehman 2020-12-10
i get feed back from people that know better than me, helps in learning. i deleted the previous one because you helped in figuring out the limited application of my code. Thank you!
please, tell me where i went wrong in this code.
i am new to MATLAB/Coding!
Rik
Rik 2020-12-10
These homework solutions are probably not the best place to get feedback. After completing the Onramp tutorial (which is provided for free by Mathworks), I would suggest looking at this thread.
And where you went wrong is in assuming what == does, instead of reading the documentation. There is an important difference between equals (which is called when you write ==) and isequal. size(A)==[row column] will result in a two-element logical vector if A is a vector or 2D array (and an error if A has more dimensions).
% Let's take a look at what if does:
A=[]; if A, disp(A),end
A=true; if A, disp(A),end
1
A=[true false]; if A, disp(A),end
A=[true true]; if A, disp(A),end
1 1
A=[false true]; if A, disp(A),end
A=[false false];if A, disp(A),end
Did you expect these results? What does this say about how your code would work?

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mohammad elyoussef
function b = freezing(a)
f = a < 32;
b = sum(f);

Yash Agarwal
Yash Agarwal 2020-4-22
function numfreeze = freezing(A)
B = A(A<32);
numfreeze = size(B,2);
end

Rajeev Mehndiratta
Rajeev Mehndiratta 2020-10-28
function numfreeze = freezing(V)
A=0
V
V(V<32) = A
A =logical(V)
B=sum(A)
numfreeze=B
end
  1 个评论
Rik
Rik 2020-10-28
You're correct, this solution didn't exist in this thread yet, although I would clean it up to this:
function numfreeze = freezing(V)
V(V<32) = 0;
V = logical(V);
numfreeze = sum(V);
end

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Mohamed
Mohamed 2023-4-25
well, you can use the size function to determine the number of the aviable elements:
function output= freezing(x)
x_1=x(x<32); % it will output any number less than 32 of the given matrix
[b,output=size(x_1); % b will equal to the unwanted elements (rows), output will equal to the number of the wanted number ( coloumbs)
%it can be vise versa if its a coloumb matrix%
  2 个评论
John D'Errico
John D'Errico 2023-4-25
This is identical to at least one other solution already posed, except for one problem. Your code will fail due to a syntax error.
DGM
DGM 2023-8-19
编辑:DGM 2023-8-19
What's more remarkable is that the answer which this most closely duplicates had already been explained to be problematic. If I recall, the deleted version would have been a more appropriate comparison, but don't trust my recollection too much. These threads are all starting to blend together.
To reiterate and address what's been said in other comments, the question does not specify the vector orientation. Any decent solution should work for any vector orientation. When I say any orientation, I mean any array with only one non-singleton dimension.
Furthermore; the question text does vaguely suggest the use of array inputs. Since there's no good reason that a solution can't be generalized to support any N-D numeric array, then it's my opinion that it should. Such a solution would also support any vector.
So what are the problems with this example? Besides the missing bracket, The misuse of size() will result in the function silently returning incorrect results for anything other than a vector oriented along dim2 or higher. The fact that this actually works for higher-dim vectors was almost certainly nothing more than an accident based on a common misunderstanding of how size() works.
x = randi([0 64],100,1); % a column vector
freezing(x) % wrong output
ans = 1
x = reshape(x,1,[]); % a row vector
freezing(x) % correct output
ans = 51
x = reshape(x,1,1,[]); % a vector on dim3
freezing(x) % output is accidentally still correct
ans = 51
x = reshape(x,10,10); % a matrix
freezing(x) % wrong output
ans = 1
function output = freezing(x)
x_1 = x(x<32);
[~, output] = size(x_1); % tilde works here since R2009b
end
You could fix the problem:
function output = freezing(x)
x_1 = x(x<32);
output = numel(x_1);
end
... but that answer has already been posted, and it can still be simplified further anyway. I should also point out that the (likely autograded) assignment clearly specifies what the output variable name should be.
function numfreeze = freezing(x)
numfreeze = nnz(x<32);
end
Note now that not only is this version simpler, it's also more generalized.

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