If I have:
t=(1:1:5)'
time=1:3:100
How do I arrange data t in each column starting from 1 until the end, with an interval of 3. Which means that the data t (1 to 5) at column 1,4,7 and so on.
I've tried:
t=[1:1:5];
nt=length(temp);
time=[1:1:100];
nti=length(time);
x=zeros(nt,nti);
temp=temp';
initiator=2;
monomer=3;
post=1:3:100;
for l=1:post
step=1;
maxstep=100;
while (step<maxstep)
step=step+3;
temp=(1:1:5)';
end
t(:,l)=t;
x=[t];
end
This only shows result X with temp at column 1. I do not know how to to arrange this data at columns that I want.
Hope someone will help me. Thank you in advance.
How many dimensions does your data have? If you already have "temp" (temperature?) and "time" as your first two dimensions and you want "t" to be the third dimension, then create a three-dimension matrix.
To extract from indexes [1 4 7 10 13 16 ... ], use (1:3:end)
To extract from indexed [2 5 8 11 14 17 ... ], use (2:3:end)
In MATLAB's colon notation, the first value is the start. Second value is increment. Third value is the end value and is inclusive.
Related
I need your help with the formulation for a calculed a field in Tableau (Tableau Prep to be accurate).
I have a field called [Code Order] which contains only a series of Odd numbers (1,3,5,7,9,..) multiple times, which means it can be (1,3,1,3,5,7,1,1,1,3,5,7,9,11).
What I need is to transform these in a normal sequence of numbers so for my example above I need as a result: (1,2,1,2,3,4,1,1,1,2,3,4,5,6)
In other words when in [Code Order] I have :
1 = 1
3 = 2
5 = 3
7 = 4
9 = 5
11 = 6
13 = 7
15 = 8
...
365 = 183
For the moment my maximum is 365, which is position 183, I would like to avoid to type 182 IF formulas if possible. ;)
Thanks in advance for your help.
CYA
Plt.K
This might turn out to be more accurate in case your Code Order series is missing any values along the way.
Example series:
Alternate Field:
Tableau Setup:
You want to use the index() calculated field. Create a new field called index. The calculation is just index().
Add [Code Order] to your row shelf and index to your label. You should see something like this.
The following calculation should do the trick
CEILING([Code Order] / 2)
I have a 29736 x 6 table, which is referred to as table_fault_test_data. It has 6 columns, with names wind_direction, wind_speed, air_temperature, air_pressure, density_hubheight and Fault_Condition respectively. What I want to do is to label the data in the Fault_Condition (last table column with either a 1 or a 0 value, depending on the values in the other columns.
I would like to do the following checks (For eg.)
If wind_direction value(column_1) is below 0.0040 and above 359.9940, label 6 th column entry corresponding to the respective row of the table as a 1, else label as 0.
Do this for the entire table. Similarly, do this check for others
like air_temperature, air_pressure and so on. I know that if-else
will be used for these checks. But, I am really confused as to how I
can do this for the whole table and add the corresponding value to
the 6 th column (Maybe using a loop or something).
Any help in this
regard would be highly appreciated. Many Thanks!
EDIT:
Further clarification: I have a 29736 x 6 table named table_fault_test_data . I want to add values to the 6 th column of table based on conditions as below:-
for i = 1:29736 % Iterating over the whole table row by row
if(1st column value <x | 1st column value > y)
% Add 0 to the Corresponding element of 6 th column i.e. table_fault_test_data(i,6)
elseif (2nd column value <x | 2nd column value > y)
% Add 0 to the Corresponding element of 6 th column i.e. table_fault_test_data(i,6)
elseif ... do this for other cases as well
else
% Add 1 to the Corresponding element of 6 th column i.e. table_fault_test_data(i,6)
This is the essence of my requirements. I hope this helps in understanding the question better.
You can use logical indexing, which is supported also for tables (for loops should be avoided, if possible). For example, suppose you want to implement the first condition, and also suppose your x and y are known; also, let us assume your table is called t
logicalIndecesFirstCondition = t{:,1} < x | t{:,2} >y
and then you could refer to the rows which verify this condition using logical indexing (please refer to logical indexing
E.g.:
t{logicalIndecesFirstCondition , 6} = t{logicalIndecesFirstCondition , 6} + 1.0;
This would add 1.0 to the 6th column, for the rows for which the logical condition is true
I currently have a 4x3500 cell array. First row is a single number, 2 row is a single string, 3rd and 4th rows are also single numbers.
Ex:
1 1 2 3 3 4 5 5 5 6
hi no ya he ........ % you get the idea
28 34 18 0 3 ......
55 2 4 42 24 .....
I would like to be able to select all columns that have a certain value in the first row. ie if I wanted '1' as the first row value, it would return
1 1
hi no
28 34
55 2
Then I would like to sort based on the 2nd row's string. ie if I wanted to have'hi', it would return:
1
hi
28
55
I have attempted to do:
variable = cellArray{:,find(cellArray{1,:} == 1)}
However I keep getting:
Error using find
Too many input arguments.
or
Error using ==
Too many input arguments.
Any help would be much appreciated! :)
{} indexing will return a comma separated list which will provide multiple outputs. When you pass this to find, it's the same as passing each element of your cell array as a separate input. This is what leads to the error about to many input arguments.
You will want to surround the comma-separated list with [] to create an array or numbers. Also, you don't need find because you can just use logical indexing to grab the columns you want. Additionally, you will want to index using () to grab the relevant rows, again to avoid the comma-separated list.
variable = cellArray(:, [cellArray{1,:}] == 1)
I have the following codes which I wish to have an output matrix Rpp of (10201,3). I run this code (which takes a bit long) then I check the matrix size of Rpp and I see (1,3), I tried so many things I couldn't find any proper way. The logic of the codes is to take the 6 values (contain 4 constant values and 2 variable values (chosen from 101 values)) and make the calculation for 3 different i1 and store every output vector of 3 in a matrix with (101*101 (pairs of those 2 variable values)) rows and 3 (for each i1) columns.
I appreciate your help
Vp1=linspace(3000,3500,101);
Vp2=3850;
rho1=2390;
rho2=2510;
Vs1=linspace(1250,1750,101);
Vs2=2000;
i1=[10 25 40];
Rpp = zeros(length(Vp1)*length(Vs1),length (i1));
for n=1:length(Vp1)*length(Vs1)
for m=1:length (i1)
for l=1:length(Vp1)
for k=1:length(Vs1)
p=sin(i1)/Vp1(l);
i2=asin(p*Vp2);
j1=asin(p*Vs1(k));
j2=asin(p*Vs2);
a=rho2*(1-2*Vs2^2*p.^2)-rho1*(1-2*Vs1(k).^2*p.^2);
b=rho2*(1-2*Vs2^2*p.^2)+2*rho1*Vs1(k)^2*p.^2;
c=rho1*(1-2*Vs1(k)^2*p.^2)+2*rho2*Vs2^2*p.^2;
d=2*(rho2*Vs2^2-rho1*Vs1(k)^2);
E=b.*cos(i1)./Vp1(l)+c.*cos(i2)/Vp2;
F=b.*cos(j1)./Vs1(k)+c.*cos(j2)/Vs2;
G=a-d*(cos(i1)/Vp1(l)).*(cos(j2)/Vs2);
H=a-d*(cos(i2)/Vp2).*(cos(j1)/Vs1(k));
D=E.*F+G.*H.*p.^2;
Rpp=((b.*(cos(i1)/Vp1(l))-c.*cos((i2)/Vp2)).*F-(a+d*((cos(i1)/Vp1(l))).*(cos(j2)/Vs2)).*H.*p.^2)./D
end
end
end
end
Try this. You 2 outer loops didn't do anything. You never used m or n so I killed those 2 loops. Also you just kept overwriting Rpp on every loop so your initialization of Rpp didn't do anything. I added an index var to assign the results to the equation to what I think is the correct part of Rpp.
Vp1=linspace(3000,3500,101);
Vp2=3850;
rho1=2390;
rho2=2510;
Vs1=linspace(1250,1750,101);
Vs2=2000;
i1=[10 25 40];
Rpp = zeros(length(Vp1)*length(Vs1),length (i1));
index = 1;
for l=1:length(Vp1)
for k=1:length(Vs1)
p=sin(i1)/Vp1(l);
i2=asin(p*Vp2);
j1=asin(p*Vs1(k));
j2=asin(p*Vs2);
a=rho2*(1-2*Vs2^2*p.^2)-rho1*(1-2*Vs1(k).^2*p.^2);
b=rho2*(1-2*Vs2^2*p.^2)+2*rho1*Vs1(k)^2*p.^2;
c=rho1*(1-2*Vs1(k)^2*p.^2)+2*rho2*Vs2^2*p.^2;
d=2*(rho2*Vs2^2-rho1*Vs1(k)^2);
E=b.*cos(i1)./Vp1(l)+c.*cos(i2)/Vp2;
F=b.*cos(j1)./Vs1(k)+c.*cos(j2)/Vs2;
G=a-d*(cos(i1)/Vp1(l)).*(cos(j2)/Vs2);
H=a-d*(cos(i2)/Vp2).*(cos(j1)/Vs1(k));
D=E.*F+G.*H.*p.^2;
Rpp(index,:)=((b.*(cos(i1)/Vp1(l))-c.*cos((i2)/Vp2)).*F-(a+d*((cos(i1)/Vp1(l))).*(cos(j2)/Vs2)).*H.*p.^2)./D;
index = index+1;
end
end
Results:
>> size(Rpp)
ans =
10201 3
The way you use the for loop is wrong. You're running the calculation for length(Vp1)*length(Vs1) * length (i1) * length(Vp1) * length(Vs1) times. Here's the correct way. I changed l into lll just so I won't confuse it with the number 1. In each iteration of the first for loop, you're running length(Vs1) times, and you need to assign the result (a 1X3 array) to the Rpp by using a row number specified by k+(lll-1)*length(Vp1).
for lll=1:length(Vp1)
for k=1:length(Vs1)
p=sin(i1)/Vp1(lll);
i2=asin(p*Vp2);
j1=asin(p*Vs1(k));
j2=asin(p*Vs2);
a=rho2*(1-2*Vs2^2*p.^2)-rho1*(1-2*Vs1(k).^2*p.^2);
b=rho2*(1-2*Vs2^2*p.^2)+2*rho1*Vs1(k)^2*p.^2;
c=rho1*(1-2*Vs1(k)^2*p.^2)+2*rho2*Vs2^2*p.^2;
d=2*(rho2*Vs2^2-rho1*Vs1(k)^2);
E=b.*cos(i1)./Vp1(lll)+c.*cos(i2)/Vp2;
F=b.*cos(j1)./Vs1(k)+c.*cos(j2)/Vs2;
G=a-d*(cos(i1)/Vp1(lll)).*(cos(j2)/Vs2);
H=a-d*(cos(i2)/Vp2).*(cos(j1)/Vs1(k));
D=E.*F+G.*H.*p.^2;
Rpp(k+(lll-1)*length(Vp1),:)=((b.*(cos(i1)/Vp1(lll))-c.*cos((i2)/Vp2)).*F-(a+d*((cos(i1)/Vp1(lll))).*(cos(j2)/Vs2)).*H.*p.^2)./D;
end
end
So, presume a matrix like so:
20 2
20 2
30 2
30 1
40 1
40 1
I want to count the number of times 1 occurs for each unique value of column 1. I could do this the long way by [sum(x(1:2,2)==1)] for each value, but I think this would be the perfect use for the UNIQUE function. How could I fix it so that I could get an output like this:
20 0
30 1
40 2
Sorry if the solution seems obvious, my grasp of loops is very poor.
Indeed unique is a good option:
u=unique(x(:,1))
res=arrayfun(#(y)length(x(x(:,1)==y & x(:,2)==1)),u)
Taking apart that last line:
arrayfun(fun,array) applies fun to each element in the array, and puts it in a new array, which it returns.
This function is the function #(y)length(x(x(:,1)==y & x(:,2)==1)) which finds the length of the portion of x where the condition x(:,1)==y & x(:,2)==1) holds (called logical indexing). So for each of the unique elements, it finds the row in X where the first is the unique element, and the second is one.
Try this (as specified in this answer):
>>> [c,~,d] = unique(a(a(:,2)==1))
c =
30
40
d =
1
3
>>> counts = accumarray(d(:),1,[],#sum)
counts =
1
2
>>> res = [c,counts]
Consider you have an array of various integers in 'array'
the tabulate function will sort the unique values and count the occurances.
table = tabulate(array)
look for your unique counts in col 2 of table.