I am having some problems regarding slicing assignment:
As i understand that general syntax of slicing is l[start:stop:step]
when we use positive step then we transverse forward and when we use negative step we transverse backward:
l=[1,2,3,4]
l[3:1:1]=[5]
when i use the above assignment then it inserts the element 5 at the index 3 like insert operation
but when i use
l[-3:-1:-1]=[5]
then it shows me value error....
i m totally confused..
please explain it.
Assuming you are asking about slices in Python,
the 'step' part will make the slice an extended slice.
Assigning to extended slices is only possible if the list on the right
hand side is of the same size as the extended slice.
see
https://docs.python.org/2.3/whatsnew/section-slices.html
So the confusing thing actually is that your l[3:1:1] = [5] does not raise
a ValueError, because the left and right size differ (0 and 1; note
that both your l[3:1:1] and l[-3:-1:-1] evaluate to empty lists).
I think that can be explained by the fact that a step of 1 is no different
from the original slice syntax [start:end], and may therefore be handled
as a normal slice.
If your goal is inserting, just don't use the step.
Related
I've come across some unexpected behaviour in matlab that I can't make sense of when performing vectorised assignment:
>> q=4;
>> q(q==[1,3,4,5,7,8])
The logical indices contain a true value outside of the array bounds.
>> q(q==[1,3,4,5,7,8])=1
q =
4 0 1
Why does the command q(q==[1,3,4,5,7,8]) result in an error, but the command q(q==[1,3,4,5,7,8])=1 work? And how does it arrive at 4 0 1 being the output?
The difference between q(i) and q(i)=a is that the former must produce the value of an array element; if i is out of bounds, MATLAB chooses to give an error rather than invent a value (good choice IMO). And the latter must write a value to an array element; if i is out of bounds, MATLAB chooses to extend the array so that it is large enough to be able to write to that location (this has also proven to be a good choice, it is useful and used extensively in code). Numeric arrays are extended by adding zeros.
In your specific case, q==[1,3,4,5,7,8] is the logical array [0,0,1,0,0,0]. This means that you are trying to index i=3. Since q has a single value, reading at index 3 is out of bounds, but we can write there. q is padded to size 3 by adding zeros, and then the value 1 is written to the third element.
I am working on a code in Least Square Non Negative solution recovery context on Matlab, and I need (with no more details because it's not that important for this question) to know the number of non zero elements in my matrices and arrays.
The function NNZ on matlab does exactly what I want, but it happens that I need more information about what Matlab thinks of a "zero element", it could be 0 itself, or the numerical zero like 1e-16 or less.
Does anybody has this information about the NNZ function, cause I couldn't get the original script
Thanks.
PS : I am not an expert on Matlab, so accept my apologies if it's a really simple task.
I tried "open nnz", on Matlab but I only get a small script of commented code lines...
Since nnz counts everything that isn't an exact zero (i.e. 1e-100 is non-zero), you just have to apply a relational operator to your data first to find how many values exceed some tolerance around zero. For a matrix A:
n = nnz(abs(A) > 1e-16);
Also, this discussion of floating-point comparison might be of interest to you.
You can add in a tolerance by doing something like:
nnz(abs(myarray)>tol);
This will create a binary array that is 1 when abs(myarray)>tol and 0 otherwise and then count the number of non-zero entries.
The following error occurs quite frequently:
Subscript indices must either be real positive integers or logicals
I have found many questions about this but not one with a really generic answer. Hence I would like to have the general solution for dealing with this problem.
Subscript indices must either be real positive integers or logicals
In nearly all cases this error is caused by one of two reasons. Fortunately there is an easy check for this.
First of all make sure you are at the line where the error occurs, this can usually be achieved by using dbstop if error before you run your function or script. Now we can check for the first problem:
1. Somewhere an invalid index is used to access a variable
Find every variable, and see how they are being indexed. A variable being indexed is typically in one of these forms:
variableName(index,index)
variableName{index,index}
variableName{indices}(indices)
Now simply look at the stuff between the brackets, and select every index. Then hit f9 to evaluate the result and check whether it is a real positive integer or logical. Visual inspection is usually sufficient (remember that acceptable values are in true,false or 1,2,3,... BUT NOT 0) , but for a large matrix you can use things like isequal(index, round(index)), or more exactly isequal(x, max(1,round(abs(x)))) to check for real positive integers. To check the class you can use class(index) which should return 'logical' if the values are all 'true' or 'false'.
Make sure to check evaluate every index, even those that look unusual as per the example below. If all indices check out, you are probably facing the second problem:
2. A function name has been overshadowed by a user defined variable
MATLAB functions often have very intuitive names. This is convenient, but sometimes results in accidentally overloading (builtin) functions, i.e. creating a variable with the same name as a function for example you could go max = 9 and for the rest of you script/function Matlab will consider max to be a variable instead of the function max so you will get this error if you try something like max([1 8 0 3 7]) because instead of return the maximum value of that vector, Matlab now assumes you are trying to index the variable max and 0 is an invalid index.
In order to check which variables you have you can look at the workspace. However if you are looking for a systematic approach here is one:
For every letter or word that is followed by brackets () and has not been confirmed to have proper indices in step 1. Check whether it is actually a variable. This can easily be done by using which.
Examples
Simple occurrence of invalid index
a = 1;
b = 2;
c = 3;
a(b/c)
Here we will evaluate b/c and find that it is not a nicely rounded number.
Complicated occurrence of invalid index
a = 1;
b = 2;
c = 3;
d = 1:10;
a(b+mean(d(cell2mat({b}):c)))
I recommend working inside out. So first evaluate the most inner variable being indexed: d. It turns out that cell2mat({b}):c, nicely evaluates to integers. Then evaluate b+mean(d(cell2mat({b}):c)) and find that we don't have an integer or logical as index to a.
Here we will evaluate b/c and find that it is not a nicely rounded number.
Overloaded a function
which mean
% some directory\filename.m
You should see something like this to actually confirm that something is a function.
a = 1:4;
b=0:0.1:1;
mean(a) = 2.5;
mean(b);
Here we see that mean has accidentally been assigned to. Now we get:
which mean
% mean is a variable.
In Matlab (and most other programming languages) the multiplication sign must always be written. While in your math class you probably learned that you can write write a(a+a) instead of a*(a+a), this is not the same in matlab. The first is an indexing or function call, while the second is a multiplication.
>> a=0
a =
0
>> a*(a+a)
ans =
0
>> a(a+a)
Subscript indices must either be real
positive integers or logicals.
Answers to this question so far focused on the sources of this error, which is great. But it is important to understand the powerful yet very intuitive feature of matrix indexing in Matlab. Hence how indexing works and what is a valid index would help avoid this error in the first place by using valid indices.
At its core, given an array A of length n, there are two ways of indexing it.
Linear indexing: with subset of integers from 1 : n (duplicates allowed). 0 is not allowed, as Matlab arrays are 1-based, unless you use the method below. For higher-dimensional arrays, multiple subscripts are internally converted into a linear index, although in an efficient and transparent manner.
Logical indexing:wherein you use a n-length array of 0s and 1s, to pick those elements where indexing is true. In this case, unique(index) must have only 0 and 1.
So a valid indexing array into another array with n number of elements ca be:
entirely logical of the same size, or
linear with subsets of integers from 1:n
Keeping this in mind, invalid indexing error occurs when you mix the two types of indexing: one or more zeros occur in your linearly indexing array, or you mix 0s and 1s with anything other than 0s and 1s :)
There is tons of material online to learn this including this one:
http://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html
The following error occurs quite frequently:
Subscript indices must either be real positive integers or logicals
I have found many questions about this but not one with a really generic answer. Hence I would like to have the general solution for dealing with this problem.
Subscript indices must either be real positive integers or logicals
In nearly all cases this error is caused by one of two reasons. Fortunately there is an easy check for this.
First of all make sure you are at the line where the error occurs, this can usually be achieved by using dbstop if error before you run your function or script. Now we can check for the first problem:
1. Somewhere an invalid index is used to access a variable
Find every variable, and see how they are being indexed. A variable being indexed is typically in one of these forms:
variableName(index,index)
variableName{index,index}
variableName{indices}(indices)
Now simply look at the stuff between the brackets, and select every index. Then hit f9 to evaluate the result and check whether it is a real positive integer or logical. Visual inspection is usually sufficient (remember that acceptable values are in true,false or 1,2,3,... BUT NOT 0) , but for a large matrix you can use things like isequal(index, round(index)), or more exactly isequal(x, max(1,round(abs(x)))) to check for real positive integers. To check the class you can use class(index) which should return 'logical' if the values are all 'true' or 'false'.
Make sure to check evaluate every index, even those that look unusual as per the example below. If all indices check out, you are probably facing the second problem:
2. A function name has been overshadowed by a user defined variable
MATLAB functions often have very intuitive names. This is convenient, but sometimes results in accidentally overloading (builtin) functions, i.e. creating a variable with the same name as a function for example you could go max = 9 and for the rest of you script/function Matlab will consider max to be a variable instead of the function max so you will get this error if you try something like max([1 8 0 3 7]) because instead of return the maximum value of that vector, Matlab now assumes you are trying to index the variable max and 0 is an invalid index.
In order to check which variables you have you can look at the workspace. However if you are looking for a systematic approach here is one:
For every letter or word that is followed by brackets () and has not been confirmed to have proper indices in step 1. Check whether it is actually a variable. This can easily be done by using which.
Examples
Simple occurrence of invalid index
a = 1;
b = 2;
c = 3;
a(b/c)
Here we will evaluate b/c and find that it is not a nicely rounded number.
Complicated occurrence of invalid index
a = 1;
b = 2;
c = 3;
d = 1:10;
a(b+mean(d(cell2mat({b}):c)))
I recommend working inside out. So first evaluate the most inner variable being indexed: d. It turns out that cell2mat({b}):c, nicely evaluates to integers. Then evaluate b+mean(d(cell2mat({b}):c)) and find that we don't have an integer or logical as index to a.
Here we will evaluate b/c and find that it is not a nicely rounded number.
Overloaded a function
which mean
% some directory\filename.m
You should see something like this to actually confirm that something is a function.
a = 1:4;
b=0:0.1:1;
mean(a) = 2.5;
mean(b);
Here we see that mean has accidentally been assigned to. Now we get:
which mean
% mean is a variable.
In Matlab (and most other programming languages) the multiplication sign must always be written. While in your math class you probably learned that you can write write a(a+a) instead of a*(a+a), this is not the same in matlab. The first is an indexing or function call, while the second is a multiplication.
>> a=0
a =
0
>> a*(a+a)
ans =
0
>> a(a+a)
Subscript indices must either be real
positive integers or logicals.
Answers to this question so far focused on the sources of this error, which is great. But it is important to understand the powerful yet very intuitive feature of matrix indexing in Matlab. Hence how indexing works and what is a valid index would help avoid this error in the first place by using valid indices.
At its core, given an array A of length n, there are two ways of indexing it.
Linear indexing: with subset of integers from 1 : n (duplicates allowed). 0 is not allowed, as Matlab arrays are 1-based, unless you use the method below. For higher-dimensional arrays, multiple subscripts are internally converted into a linear index, although in an efficient and transparent manner.
Logical indexing:wherein you use a n-length array of 0s and 1s, to pick those elements where indexing is true. In this case, unique(index) must have only 0 and 1.
So a valid indexing array into another array with n number of elements ca be:
entirely logical of the same size, or
linear with subsets of integers from 1:n
Keeping this in mind, invalid indexing error occurs when you mix the two types of indexing: one or more zeros occur in your linearly indexing array, or you mix 0s and 1s with anything other than 0s and 1s :)
There is tons of material online to learn this including this one:
http://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html
When I try to do the following command I get an error.
err = sqrt(mean((xi256-xc256).^2))
I am aware that the matrix sizes are different.
whos xi256 xc256` gives:
Name Size Bytes Class Attributes
xc256 27x1 216 double
xi256 513x1 4104 double
I am supposed to negate find the difference of these two matrices. In fact the command given at the top was in the course notes and the course has been running for several years! I have tried googling ways to resolve this error to perform this subtraction regardless but have found no solution. Maybe one of the matrices can be scaled to match the dimensions of the other? However, I have not been able to find any such functions that would let me do this.
I need to find the RMS error in a given set of data. xc256 was calculated through a numerical method, xi256 gives the true value.
Edit: I was able to use another set of results.
First check that xc256 is correctly computed and that you do not have a matrix transposition gone wrong or something like that. Computing the RMS between two vector of different sizes does not make sense, and padding or replicating will get you rid of the error, but is most probably not what you actually want.
There are just two situations that I can think of, I will list them here:
The line is wrong (not likely as it looks pretty normal, but make sure to check the book!)
The input of the line is wrong
Assuming it is in point 2, again there are two possibilities:
xi256 is of incorrect size (likelyhood of this depends on how you got it)
xc256 is of incorrect size
Assuming it is again point 2, there are yet again 2 likely possibilities:
xc256 should be a scalar
xc256 should be a vector with the same size as xi256
From here on there is insufficient information to continue, but check whether you accidentally made it 27 times as long, or 19 times too short somewhere. Just use some breakpoints throughout the code to see how the size develops.