Say you have an array, data, of unknown length. Is there a shorter method to get elements form a starting index to the end than
subdata = data(2:length(data))
You can use end notation to indicate the last element. data(2:end) returns a vector containing elements in the vector data from element 2 to the last element. Or if data is a character array, it returns the second character all the way to the last character. And data(end) returns the last element.
This can be done with matrices too, i.e. data(2:end,5:end). Additionally you can use it as an operand, i.e. data(2:end-1) , data(2:end/2).
In this context, end serves a different purpose from its use at the end of functions/loops/switches.
Related
Write a C program to determine the maximum value among the elements of the array, the first and last position of the element with this value.
I tried to determine the position of the first maximal element, but failed.
I struggle with printing text in my Matlab GUI.
I have code like this in my callback:
if Lia == ismember(handles.T(1:3),(1,1,1))
set(handles.t1, 'String', 'good day');
end
The problem is, I don't know how to check if in my array indexes from 1 to 3 I got this numbers: 1,1,1. I was looking to the documentation but it appears it says nothing about that (or I simply cannot find the proper answer).
You can simply use all and check to see if every element in the first three slots of your array match the values of 1 explicitly. I don't know the shape of your array so I'm going to force it to be a column vector. If the first three slots of the array was a row or column vector and if we assumed that the values of 1 are a column or row vector respectively then you're going to get a rather unpleasant surprise:
h = handles.T(1:3);
if all(h(:) == [1; 1; 1])
set(handles.t1, 'String', 'good day');
end
Note that I could have simply done all(h(:) == 1) as a special case since we are performing a comparison of every element in an array with a single value. However, I have a feeling that this may change for you, so I've decided to explicitly make a vector of 1s so you can change the contents of what you want to compare to at a later time.
I would like to know what the meaning of
col2=b1(1:end,lead);
is in MATLAB?
This is very basic stuff. We are led to assume that b1 is a 2-dimensional array. The contents of an array are indexed using brackets, so b1(1,1) will return the top left element of array b1. The first index in your example, 1:end is selecting every element in the first dimension (matlab indexes rows first, then columns). The second index, lead in your example, selects a particular column. We assume that lead has been allocated previously somewhere in the code.
Thence, b1(1:end,lead) returns to col2 a 1-dimensional array containing a subarray of b1. This could be accomplished more elegantly using col2=b1(:,lead);.
I'm trying to initialize
labels =['dh';'Dh';'gj';'Gj';'ll';'Ll';'nj';'Nj';'rr';'Rr';'sh';'Sh';'th';'Th';'xh';'Xh';'zh';'Zh';'ç';'Ç';'ë';'Ë'];
But it shows me the error on title.When I try with numbers it's all perfect but not with characters.What could be the problem?
If you wish to eliminate any padding, you can also store it into a cell as follows.
labels = {'dh';'Dh';'gj';'Gj';
'll';'Ll';'nj';'Nj';
'rr';'Rr';'sh';'Sh';
'th';'Th';'xh';'Xh';
'zh';'Zh';'ç';'Ç';
'ë';'Ë'};
Then you can reference the "i"th element using labels{i} instead of labels(i,:) which is simpler. You can further run more string operations using cellfun and not interfere with any existing values that you've stored.
I agree with krisdestruction that using a cell array makes the code accessing the strings simpler and is generally more idiomatic. That is what I would also recommend unless there is a compelling reason to do something else.
For completeness, you could use the char function to add the padding automatically for you if you really want a character array:
>> char('aa','bb','c')
ans =
aa
bb
c
where the last row is 'c '. From the char documentation:
S = char(A1,...,AN) converts the arrays A1,...,AN into a single character array. After conversion to characters, the input arrays become rows in S. Each row is automatically padded with blanks as needed. An empty string becomes a row of blanks.
(Emphasis mine)
From the Mathworks documentation:
Apply the MATLAB concatenation operator, []. Separate each row with a semicolon (;). Each row must contain the same number of characters. For example, combine three strings of equal length:
You can try padding like this to make every row 2 characters:
labels = ['dh';'Dh';'gj';'Gj';
'll';'Ll';'nj';'Nj';
'rr';'Rr';'sh';'Sh';
'th';'Th';'xh';'Xh';
'zh';'Zh';'ç ';'Ç ';
'ë ';'Ë '];
Part 1
I know that QuickSort can be used 'in place' but could someone explain to me how Insertion sort Algorithm does this using 'in place'.
From my understanding:
Insertion Sort starts at the first value and compares it to the next value, if that value is less than our value they switch positions. We continue this recursively. (Short explanation)
So would you say that this is 'in place' because we don't create a new array to do this, but just compare two elements in an array?
If my understanding was wrong could someone please explain the
algorithm for insertion sort in place.
Part 2
Also how would I use insertion sort to illustrate the idea of a loop invariant?
I know that a loop invariant is a condition that is immediately true before and after each iteration of a loop but I'm not sure how this would relate to an insertion sort.
Like Thorsten mentioned in the comments section, you have described bubble sort. Modified from Wikipedia, pseudocode for bubble sort is as follows:
procedure bubbleSort( A : list of sortable items )
n = length(A)
for i = 1 to n inclusive do // Outer Loop
for j = 1 to n-1-i inclusive do
/* if this pair is out of order */
if A[j] > A[j+1] then
swap(A[j], A[j+1])
end if
end for
end for
end procedure
The loop invariant in bubble sort (for the outer loop) would be that, after each iteration of the loop, the entire array until the current value of i would be sorted. This is because, each time one reaches the outer loop, it will be after going through all iterations of the inner loop (from i to n-1), finding the minimum element there and swapping that element with the ith one.
Bubble sort is, indeed, in place, since all the sorting happens within the original array itself, and does not require a separate external array.
Edit- now onto insertion sort:
Pseudo code is as follows (all hail Wikipedia):
for i = 1 to length(A) - 1
x = A[i]
j = i
while j > 0 and A[j-1] > x
A[j] = A[j-1]
j = j - 1
end while
A[j] = x[3]
end for
Here, at each step, what happens is that for each element, you select the appropriate location at which to insert it into the array, i.e., you insert it just after the first element that is smaller than it in the array. In a little more detail, what the inner loop does is that, it keeps shifting elements to the right till it encounter an element smaller than the element in consideration, at which point you insert the element just after the smaller element. What this will mean is that every element until the aforementioned element is sorted. the outer loop ensures that this is done for all elements within the array, which means that by the time the outer loop completes, the array is sorted.
The loop invariant for the outer loop is, like before, that after the ith iteration, all elements till the current i will be sorted. Just before the ith interation, however, all elements till i-1 will be sorted.
The insertion sort algorithm does not require an external array for sorting. More specifically, all operations are done on the array itself (except for the one variable we need to store the element that we are currently trying to insert into its appropriate location), and no external arrays are used- there is no copying from this array to another one, for example. So, the space complexity required by the algorithm (excluding, of course, the space the array itself occupies) will be O(1), as opposed to dependent on the size of the array, and the sort is in-place, much like bubble sort.