Consider the following example code:
spacing_Pin = transpose(-27:0.001:2);
Phi_intrp3 = interp1(P_in3,Phi_out_deg3,spacing_Pin,'spline');
size(P_in3) = size(Phi_out_deg3) = 40 1
P_in1st = P_in3 -0.76;
thetah_1st = Phi_intrp3(ismember(spacing_Pin,P_in1st));
I think thetah_1stshould now have 40 elements. It turns out, however, it only has 20. This must be, because P_in1stcan only be found 20 times in spacing_Pin. I know, you're not given P_in3, but it only has up to 2 digits after the dot. So should P_in1st. min(P_in3) = -25.9800and max(P_in3) = -0.0200. These values should be included in spacing_Pin. Why does thetah_1st have only 20 entries.
It's impossible to say without more information and your actual variable values, but I suspect that the issue is with using ismember the way that you have.
Since you are looking for floating point numbers within an array rather than integers, you should use ismembertol (ismember with a tolerance) to handle any potential floating point errors.
thetah_1st = Phi_intrp3(ismembertol(spacing_Pin, P_in1st));
Related
I know how to 'select' a range in LO (7.2.4.1) Calc BASIC ....
ThisComponent.CurrentController.ActiveSheet.getCellRangeByName("D1:H6")
But how to write a value, e.g. "1", into that range using BASIC?
myRange = ThisComponent.CurrentController.ActiveSheet.getCellRangeByName("D1:H6")
myRange.Value = 1
Gives an "property or method not found" error. But I can't find any properties or values to go after Range to allow me to do what I want. Flailing around and trying
myRange.setValue = 1
myRange.writeValue = 1
myRange.setString = "1"
and numerous other variants don't work either.
Would really appreciate the solution. Thanks.
You can edit the value of an individual cell, but not the entire range. You will have to iterate over all the cells in the range one at a time, changing the value of each of them.
Sub Set1ToD1H6
myRange = ThisComponent.CurrentController.ActiveSheet.getCellRangeByName("D1:H6")
For i = 0 To myRange.getRows().getCount()-1
For j = 0 To myRange.getColumns().getCount()-1
myRange.getCellByPosition(j, i).setValue(1)
Next j
Next i
End Sub
But since the read-write operation to a cell is comparable in time to the read-write operation to a whole range, it is preferable to use another method - to prepare data in an array and write from it to a range in one operation:
Sub Set1ToRange
myRange = ThisComponent.CurrentController.ActiveSheet.getCellRangeByName("D1:H6")
dataOfRange = myRange.getData()
For i = LBound(dataOfRange) To UBound(dataOfRange)
For j = LBound(dataOfRange(i)) To UBound(dataOfRange(i))
dataOfRange(i)(j) = 1
Next j
Next i
myRange.setData(dataOfRange)
End Sub
(For your example, this will be approximately 30 times faster, for a larger range the time winnings will be even more significant)
The .getData() and .setData() methods work on numeric range values. To work with text strings (and numbers), use .getDataArray() and .setDataArray(), for working with cell formulas use .getFormulaArray() and .setFormulaArray()
I have an array of 1 x 400, where all element values are above 1500. However, I have some elements that have values<50 which are wrong measures and I would like to have the mean of the elements before and after the wrong measured data points and replace it in the main array.
For instance, element number 17 is below 50 so I want to take the mean of elements 16 and 18 and replace element 17 with the new mean.
Can someone help me, please? many thanks in advance.
No language is specified in the question, but for Python you could work with List Comprehension:
# array with 400 values, some of which are incorrect
arr = [...]
arr = [arr[i] if arr[i] >= 50 else (arr[i-1]+arr[i+1])/2 for i in range(len(arr))]
That is, if arr[i] is less than 50, it'll be replaced by the average value of the element before and after it. There are two issues with this approach.
If i is the first or last element, then one of the two values will be undefined, and no mean can be obtained. This can be fixed by just using the value of the available neighbour, as specified below
If two values in a row are very low, the leftmost one will use the rightmost one to calculate its value, which will result in a very low value. This is a problem that may not occur for you in practice, but it is an inherent result of the way you wish to recalculate values, and you might want to keep it in mind.
Improved version, keeping in mind the edge cases:
# don't alter the first and last item, even if they're low
arr = [arr[i] if arr[i] >= 50 or i == 0 or i+1 == len(arr) else (arr[i-1]+arr[i+1])/2 for i in range(len(arr))]
# replace the first and last element if needed
if arr[0] < 50:
arr[0] = arr[1]
if arr[len(arr)-1] < 50:
arr[len(arr)-1] = arr[len(arr)-2]
I hope this answer was useful for you, even if you intend to use another language or framework than python.
As simple as in title. I have nx1 sized vector p. I'm interested in the maximum value of r = p/foo - floor(p/foo), with foo being a scalar, so I just call:
max_value = max(p/foo-floor(p/foo))
How can I get which value of p gave out max_value?
I thought about calling:
[max_value, max_index] = max(p/foo-floor(p/foo))
but soon I realised that max_index is pretty useless. I'm sorry asking this, real beginner here.
Having dropped the issue to pieces, I realized there's no unique corrispondence between values p and values in my related vector p/foo-floor(p/foo), so there's a logical issue rather than a language one.
However, given my input data, I know that the solution is unique. How can I fix this?
I ended up doing:
result = p(p/foo-floor(p/foo) == max(p/foo-floor(p/foo)))
Looks terrible, so if you know any other way...
Once you have the index, use it:
result = p(max_index)
You can create a new vector with your lets say "transformed" values:
p2 = (p/foo-floor(p/foo))
and then just use find to find the max values on p2:
max_index = find(p2 == max(p2))
that will return the index or indices of p2 with the max value of that operation, and finally just lookup the original value in p
p(max_index)
in 1 line, this is:
p(find((p/foo-floor(p/foo) == max((p/foo-floor(p/foo))))))
which is basically the same thing you did in the end :)
In MatLab, I have several data vectors that are in text. For example:
speciesname = [species1 species2 species3];
genomelength = [8 10 5];
gonometype = [RNA DNA RNA];
I realise that to make a plot, arrays must be numerical. Is there a quick and easy way to assign unique entries in an array a number, for example so that RNA = 1 and DNA = 2? Note that some arrays might not be binary (i.e. have more than two options).
Thanks!
So there is a quick way to do it, but im not sure that your plots will be very intelligible if you use numbers instead of words.
You can make a unique array like this:
u = unique(gonometype);
and make a corresponding number array is just 1:length(u)
then when you go through your data the number of the current word will be:
find(u == current_name);
For your particular case you will need to utilize cells:
gonometype = {'RNA', 'DNA', 'RNA'};
u = unique(gonometype)
u =
'DNA' 'RNA'
current = 'RNA';
find(strcmp(u, current))
ans =
2
When the numbers are really small, Matlab automatically shows them formatted in Scientific Notation.
Example:
A = rand(3) / 10000000000000000;
A =
1.0e-016 *
0.6340 0.1077 0.6477
0.3012 0.7984 0.0551
0.5830 0.8751 0.9386
Is there some in-built function which returns the exponent? Something like: getExponent(A) = -16?
I know this is sort of a stupid question, but I need to check hundreds of matrices and I can't seem to figure it out.
Thank you for your help.
Basic math can tell you that:
floor(log10(N))
The log base 10 of a number tells you approximately how many digits before the decimal are in that number.
For instance, 99987123459823754 is 9.998E+016
log10(99987123459823754) is 16.9999441, the floor of which is 16 - which can basically tell you "the exponent in scientific notation is 16, very close to being 17".
Floor always rounds down, so you don't need to worry about small exponents:
0.000000000003754 = 3.754E-012
log10(0.000000000003754) = -11.425
floor(log10(0.000000000003754)) = -12
You can use log10(A). The exponent used to print out will be the largest magnitude exponent in A. If you only care about small numbers (< 1), you can use
min(floor(log10(A)))
but if it is possible for them to be large too, you'd want something like:
a = log10(A);
[v i] = max(ceil(abs(a)));
exponent = v * sign(a(i));
this finds the maximum absolute exponent, and returns that. So if A = [1e-6 1e20], it will return 20.
I'm actually not sure quite how Matlab decides what exponent to use when printing out. Obviously, if A is close to 1 (e.g. A = [100, 203]) then it won't use an exponent at all but this solution will return 2. You'd have to play around with it a bit to work out exactly what the rules for printing matrices are.