This question already has answers here:
How does one declare an array in VBScript?
(2 answers)
Closed 5 years ago.
how to create array and display required array value. For example:
Arr[0]="anand" Arr[1]="bala".print Arr optin:2, Bala .
i need the required array output
Queat seems a bit unclear.
May be below piece of code can be relevent to your Question.
'Initializing & Filling up array
Dim testArr(3)
testArr(0)="First"
testArr(1)="Second"
testArr(2)="Third"
Dim Iterator,userInput
'If you Need ro validate the User input you need to use Err Object.
'Like if it's number or not/Is teh User Provided value is positive or not/ etc. etc. which i'm not doing now)
userInput=Cint(inputbox("Please enter a Positive number between 1-" & Ubound(testArr)))
If userInput<=Ubound(testArr) Then
MsgBox(testArr(userInput-1))
Else
MsgBox("Wrong Value Provided. Not Applicable Array Index Number !!!")
End If
Related
This question already has answers here:
Getting the indices of several elements in a NumPy array at once
(5 answers)
Closed 2 years ago.
Assume I have an array:
a = np.array([1,2,3,4,5])
Now I want to find the indices of elements in this array corresponding to the values given by another array input:
input = np.array([2,4,5])
The expected result should be:
result = [1,3,4]
A boolean mask, which is true for element indices 1,3,4 would also be fine.
I do not want to use looping to solve this. I assume that a possible solution has to do with the numpy where() function, but using this one, I am only able to compare the entries of array a with one element of array input at a time. Because the length of input might differ, I cannot really use this approach. Do you have any other ideas?
Thanks in advance.
np.where(np.in1d(a, inp))[0]
or:
np.isin(a, inp).nonzero()[0]
or as suggested here:
sorter = np.argsort(a)
sorter[np.searchsorted(a, inp, sorter=sorter)]
output:
[1 3 4]
np.where(np.in1d(a, inp))[0] np.where(np.in1d(a, inp))[0]
This question already has answers here:
Get a number from an array of digits
(3 answers)
Closed 3 years ago.
I really like Julia's Digits function which will return an array of the individual digits that make up the input integer (See code example). My question is, once you have this array, is there an easy way to join the individual values back into the original value passed into digits?
Now, I suppose I could just take each integer's index value and multiply by it's corresponding power of 10 and modify the array in place. Then I could use sum on the array, but is there a better way to do it?
I.E
function getDigits(x)
return digits(x)
end
Julia> getDigits(1234)
4-element Array{Int64,1}:
4
3
2
1
function joinDigits(digitArray)
for i in 0:length(digitArray)-1
digitArray[i+1] = digitArray[i+1] * 10 ^ i
end
return sum(digitArray)
Julia> joinDigits([4,3,2,1])
1234
foldr((rest, msd) -> rest + 10*msd,x) is a one liner that does the same thing. It's terse, but probably not as clean as the for loop.
This question already has answers here:
On shape-agnostic slicing of ndarrays
(2 answers)
Closed 6 years ago.
Suppose A is multi-dimensional array (MDA) of size 3,4,5 and B is another MDA of size 3,4,5,6.
I know A(1,:,:) or B(1,:,:,:) can both extract their elements along the first dimension.
I now need to write a general program to extract the k-th dimension from a MDA without knowing its size.
For example, the MDA C has 6 dimension: 4,5,6,7,8,9 and I want an extraction C(:,:,k,:,:,:).
Sometimes, the MDA 'D' has 4 dimension: 3,4,5,6 and I want another extraction D(k,:,:,:).
That is, my problem is the numbers of colon is varying because of the dimension.
Thanks in advance
You can use string arrays to index the array dynamically:
function out = extract(arr,dim,k)
subses = repmat({':'}, [1 ndims(arr)]);
subses(dim) = num2cell(k);
out = arr(subses{:});
where dim is the dimension in which you want to select and k is an index within that dimension.
I have used a code from this answer:
https://stackoverflow.com/a/27975910/3399825
This question already has answers here:
Remove matched item from array of objects?
(5 answers)
Closed 7 years ago.
So I have an array of Int's and I am trying to create a function to delete a certain value from the array. However, the removeAtIndex() function only deletes the first occurrence and the removeLast() only removes the last. I tried enumerating through the array but I end up with an Array index out of range error, possible due to the reshifting that occurs when deleting an item from the array.
for (index, value) in connectionTypeIDs.enumerate() {
if (value == connectionTypeToDelete){
connectionTypeIDs.removeAtIndex(index)
}
}
Any ideas on how to accomplish this?
The quickest way is to use filter. I don't know is that answer your question but you can have a look on this:
// remove 1 from array
arr = arr.filter{$0 != 1}
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Determining the number of occurrences of each unique element in a vector
I've the following array:
v = [ 1 5 1 6 7 1 5 5 1 1]
And I need to count the values and show the number that has more appearances.
From the example on the top, the solution would be 1 (there are five 1's)
Thanks in advance
Use mode.
If you need to return the number of elements as well, do the following:
m = mode(v);
n = sum(v==m);
fprintf('%d appears %d times\n',m,n);
Another method is using the hist function, if you're dealing with integers.
numbers=unique(v); %#provides sorted unique list of elements
count=hist(v,numbers); %#provides a count of each element's occurrence
Just make sure you specify an output value for the hist function, or you'll end up with a bar graph.
#Jacob is right: mode(v) will give you the answer you need.
I just wanted to add a nice way to represent the frequencies of each value:
bar(accumarray(v', 1))
will show a nice bar diagram with the count of each value in v.