Why are these zeros adding themselves to the array?

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Andrew
Andrew 2025-2-6
评论： Voss 2025-2-6
Ran in:
clear
clc
function result = expn(x,n)
result = 1;
for i = 1:n
result = (result + x.^i/factorial(i));
end
end
x = [1,-2];
n = [1,2,4,6,10];
ex1 = zeros(1,5);
ex_2 = zeros(1,5);
for k = n
t = expn(x(1),k);
ex1(k) = t;
end
for k = n
t = expn(x(2),k);
ex_2(k) = t;
end
ex1
ex1 = 1×10
2.0000 2.5000 0 2.7083 0 2.7181 0 0 0 2.7183
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ex_2
ex_2 = 1×10
-1.0000 1.0000 0 0.3333 0 0.1556 0 0 0 0.1354
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Not sure why the zeros are appearing as elements in the arrays. If someone could help me out it would be much appreciated!
  2 个评论
Torsten
Torsten 2025-2-6
编辑：Torsten 2025-2-6
The zeros are at the positions that are not covered by n = [1,2,4,6,10];. Would you prefer a number different from 0 to be set there ?
Andrew
Andrew 2025-2-6
Oh that makes sense. No I'd rather there be no zeros, and have the array consist only of the nonzero numbers.

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采纳的回答

Voss
Voss 2025-2-6
编辑：Voss 2025-2-6
Ran in:
k goes 1,2,4,6,10
n = [1,2,4,6,10];
for k = n
fprintf('k = %d\n',k);
end
k = 1 k = 2 k = 4 k = 6 k = 10
so inside your loop, e.g., ex1(k) = t; sets ex1(1), ex1(2), ex1(4), ex1(6), ex1(10)
When you set an element of an array that's outside the current size of the array, the array is expanded as necessary with elements containing zeros. So that's where the zeros are coming from.
Example:
vec = zeros(1,5)
vec = 1×5
0 0 0 0 0
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vec(4) = 40 % not expanded
vec = 1×5
0 0 0 40 0
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vec(10) = 100 % vec gets expanded to length 10 with zeros
vec = 1×10
0 0 0 40 0 0 0 0 0 100
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You probably meant for ex1 and ex_2 to be length-5 vectors for their entire lifetimes, in which case you'd do something like this:
x = [1,-2];
n = [1,2,4,6,10];
ex1 = zeros(1,5);
ex_2 = zeros(1,5);
m = numel(n);
for k = 1:m
t = expn(x(1),n(k));
ex1(k) = t;
end
for k = 1:m
t = expn(x(2),n(k));
ex_2(k) = t;
end

更多回答（2 个）

Walter Roberson
Walter Roberson 2025-2-6
Ran in:
function result = expn(x,n)
result = 1;
for i = 1:n
result = (result + x.^i/factorial(i));
end
end
x = [1,-2];
n = [1,2,4,6,10];
ex1 = dictionary();
ex_2 = dictionary;
for k = n
t = expn(x(1),k);
ex1(k) = t;
end
for k = n
t = expn(x(2),k);
ex_2(k) = t;
end
ex1
ex1 = dictionary (double --> double) with 5 entries: 1 --> 2 2 --> 2.5000 4 --> 2.7083 6 --> 2.7181 10 --> 2.7183
ex_2
ex_2 = dictionary (double --> double) with 5 entries: 1 --> -1 2 --> 1 4 --> 0.3333 6 --> 0.1556 10 --> 0.1354
Catalytic
Catalytic 2025-2-6
Ran in:
ex1=[2.0000 2.5000 0 2.7083 0 2.7181 0 0 0 2.7183]
ex1 = 1×10
2.0000 2.5000 0 2.7083 0 2.7181 0 0 0 2.7183
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<mw-icon class=""></mw-icon>
ex1=nonzeros(ex1)'
ex1 = 1×5
2.0000 2.5000 2.7083 2.7181 2.7183
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<mw-icon class=""></mw-icon>

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