I have a relatively large data set, and I'm looking for the missing number via MatLab.
For example, I have a list of numbers that might look like:
1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 5, 5, 6, 6, 7, 7, 7, 7, 9, 10, 10.....
You can see the 8 is missing here. The list is in the thousands, and there are maybe just a couple missing numbers. How can I find out which ones are missing? My search only turned up useful results without randomly repeating numbers. Seems simple but I can't figure it out.
Thanks for help!
Use unique, like this:
B=unique(A); % A is your data
C=setdiff(1:max(A),B)
and C is your desired missing numbers.
EDIT (afetr seeing claj's answer):
If your data starts from another value (not "1"), the second line should be:
C=setdiff(min(A):max(A),B)
EDIT2: (according to Eitan's comment)
C=setdiff(min(A):max(A),A);
This line replaces the two lines from the original answer.
You could do something like this:
% Your data:
data = [1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 5, 5, 6, 6, 7, 7, 7, 7, 9, 10, 10];
for i = 1:data(end)
if (isempty(find(data==i)))
disp(['i = ',num2str(i)]);
end
end
Which will print out the values of the missing elements.
Or even simpler you could just use the ismember() function to construct
the set difference in just a single line below.
% First enter your data and construct 'set':
data = [1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 5, 5, 6, 6, 7, 7, 7, 7, 9, 10, 10];
set = data(1):data(end);
Then to determine which elements of 'set' are also in 'data':
ismember(set, data)
The output then shows the locations in 'set' where the data is missing:
ans =
1 1 1 1 1 1 1 0 1 1
Use the ismember() function to check if a number is member of the data array
% set your data array
maximum = max(data);
minimum = min(data);
for i= minimum:maximum
if ~ismember(i,data);
disp([num2str(i) , ' is missed']);
end
end
Create a unique list of values in the array.
Find the min and max numbers in this unique set (these should be the same numbers as in the array, but quicker to find).
Create a range from min to max like [min:max].
Make a set difference of the uniqued array and the range-set.
This gives you the missing numbers in decently quick way.
this is similar to a few of the above but the simplest i've found is
find(~ismember(set,data))
which will return the indices of the members of set that are not in data
Related
I'm new to swift. can someone please explain what I'm doing wrong here.
1.
var numbers = [1, 5, 7, 6, 6, 6, 6, 6, 2]
for i in numbers{
print(numbers[i],terminator: "")
}
why doesn't this just print the numbers in the array?
2.
Here I want to set the elements in the array to a random number from 0 to 2, and then print them.
for j in numbers{
numbers[j] = Int.random(in: 0...2)
print(numbers[j],terminator: "")
}
this seems to work, but then if, outside of the for loop, I print them again:
for k in numbers{
print(numbers[k],terminator: "")
}
It outputs different numbers, from 0 to 2
3.
OK so I try a different syntax:
for m in numbers{
print(m,terminator: "")
}
now I get the same numbers every time and they are not from 0 to 2... I'm sure my mistakes are trivial but an explanation would help me out. Thanks.
to achieve what you expect, you need to loop over the indices of the array, like this:
var numbers = [1, 5, 7, 6, 6, 6, 6, 6, 2]
for i in numbers.indices { // <-- here
print(numbers[i])
}
And as mentioned, read the basics of Swift.
I would like to count the number of element in a field of a structure in Matlab
The file name is "data.m"
I have something like this:
The file has 3 fields (Columns): x, y and z
The file has maximal 6 lines, which is the number of line of the y-column
x has 5 elements (0, 9, 5, 6, 6)
y has 6 elements (6, 1, 2, 2, 8, 2)
z has 4 elements (8, 8, 4, 9)
Using:
number_of_element = numel(data.x);
returns 1.
It only takes the first element (The first line, which is "0" here)
I would like to have the number of element of the column x, which is "5" in this case.
Then I tried this:
number_of_element = numel(data(:,x));
But it doesn't work. I though Matlab could recognise "x" as a field name.
This also did not work:
count = 0;
for i = 1:end % I get an error because of this "end". Why is it not recognised here ?
number_of_element = numel(data(i).x);
count = count+number_of_element;
end
How could I get the number of element in the x-column ?
Thank you in advance.
I'm using IndexSet and I'm trying to access some indexes which at times are consecutive and at other times are not.
For example, my set may contain [1, 2, 3, 5, 6, 7, 13, 31]
I want to pull out of the set a range of 3...13, but am having difficulty with the syntax. I've learned how to use the function commands from Apple documentation, by using myIndexSet.sorted(). However, the Apple Documentation does not give an example of how to access a range of elements in the set. The Apple Documentation for accessing elements in the index set are the following:
subscript(Range<IndexSet.Index>)
I've tried a number of ways to write this but can't figure out how to do it right. Can someone show me how to access a range of elements in the set to create a new set? I've tried things such as:
let subset = subscript(Range: myLargerSet.3...13)
but it doesn't seem to work.
Thanks
What you're looking for is the intersection of your IndexSet ([1, 2, 3, 5, 6, 7, 13, 31]) with another IndexSet ([3, 4, ..., 12, 13]):
let yourIndexSet: IndexSet = [1, 2, 3, 5, 6, 7, 13, 31]
let desiredIndexRange = IndexSet(3...13)
let indicesOfInterest = yourIndexSet.intersection(desiredIndexRange)
print(indicesOfInterest.sorted()) // => [3, 5, 6, 7, 13]
One possible solution is to use a filter to create a new IndexSet.
let set = IndexSet(arrayLiteral: 1,2,3,5,6,7,13,31)
let subset = set.filteredIndexSet { (index) -> Bool in
index >= 3 && index <= 13
}
You can access a slice of your original set as follows:
let slice = indexSet[indexSet.indexRange(in: 3...13)]
slice accesses the existing elements in place, so creation of the slice is O(1)
My question is probably really easy, but I am a mathematica beginner.
I have a dataset, lets say:
Column: Numbers from 1 to 10
Column Signs
Column Other signs.
{{1,2,3,4,5,6,7,8,9,10},{d,t,4,/,g,t,w,o,p,m},{g,h,j,k,l,s,d,e,w,q}}
Now I want to extract all rows for which column 1 provides an odd number. In other words I want to create a new dataset.
I tried to work with Select and OddQ as well as with the IF function, but I have absolutely no clue how to put this orders in the right way!
Taking a stab at what you might be asking..
(table = {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ,
Characters["abcdefghij"],
Characters["ABCDEFGHIJ"]}) // MatrixForm
table[[All, 1 ;; -1 ;; 2]] // MatrixForm
or perhaps this:
Select[table, OddQ[#[[1]]] &]
{{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}}
The convention in Mathematica is the reverse of what you use in your description.
Rows are first level sublists.
Let's take your original data
mytable = {{1,2,3,4,5,6,7,8,9,10},{d,t,4,"/",g,t,w,o,p,m},{g,h,j,k,l,s,d,e,w,q}}
Just as you suggested, Select and OddQ can do what you want, but on your table, transposed. So we transpose first and back:
Transpose[Select[Transpose[mytable], OddQ[First[#]]& ]]
Another way:
Mathematica functional command MapThread can work on synchronous lists.
DeleteCases[MapThread[If[OddQ[#1], {##}] &, mytable], Null]
The inner function of MapThread gets all elements of what you call a 'row' as variables (#1, #2, etc.). So it test the first column and outputs all columns or a Null if the test fails. The enclosing DeleteCases suppresses the unmatching "rows".
Edit: made a github issue, it got closed a day later by jashkenas. So the takeaway is "working as intended" essentially.
coffee> arr
[ 0,
1,
2,
3,
'A',
'K' ]
coffee> arr[...]
[ 0,
1,
2,
3,
'A',
'K' ]
coffee> arr[..]
[ 0,
1,
2,
3,
'A',
'K' ]
According to the docs, those should be different.
With two dots (3..6), the range is inclusive (3, 4, 5, 6); with three dots (3...6), the range excludes the end (3, 4, 5).
The two slice statements that are produced are the same. Seems to me that .. should produce .slice(0) and ... should produce .slice(0, -1) Am I missing something or seeing a bug?
1.7.1
The documentation then goes on to say:
Slices indices have useful defaults. An omitted first index defaults
to zero and an omitted second index defaults to the size of the array.
This is consistent with what you're seeing. The length of your array is 6 so:
[..] is equivalent to [0..6] which would compile to .slice(0,7)
[...] is equivalent to [0...6] which would compile to .slice(0,6)
With an array of length 6, both .slice(0,6) and .slice(0,7) return all elements and so both are equivalent to .slice(0), which is what both [..] and [...] compile to.
What you are expecting would be the case if an omitted second index defaulted to the size of the array minus 1, but this is not the case.