For my task I need to convert all logical columns in a table to double. The table is sorted so I want to keep the order of columns.
Here is a mini example:
SomeString={'one';'two';'three'}
SomeNumbers={1;2;3}
SomeLogicals=[true; false; true]
T=table(SomeString, SomeNumbers, SomeLogicals)
T(:,vartype('logical'))=cell2table(cellfun(#double,table2cell(T(:,vartype('logical'))),'uniformOutput',false))
Why are the columns still logical?
T(:,vartype('logical'))
3×1 table
SomeLogicals
________
true
false
true
And how can I convert them?
Based on your reply to my comment, this should do what you want
logicalColumns = T.Properties.VariableNames(cellfun(#(x) islogical(T.(x)), T.Properties.VariableNames));
for c = 1:length(logicalColumns)
T.(logicalColumns{c}) = double(T.(logicalColumns{c}));
end
Here we get the names of the columns which contain logical values and then iterate over these columns setting them to double values of 1 or 0 for true and false.
Related
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 want to count the dates. 1 date = 1, 2 dates = 2...
I have 2 dates and I want to prepare a formula if I have 2 dates, then this is total 2.
Adjust the range references to suit your data.
=COUNTA(C2:G2)
Where C2:G2 is your first row under datum. This equation will count the number of non blank cells.
If 4.9. is a number an not text, then you could also use
=COUNT(C2:G2)
In Excel you could create a new column that checks if the cell is a date by doing =ISERROR(DAY(A1)).
If it is a date the formula will return FALSE.
Then simply count all the cells with FALSE by doing =COUNTIF(B1:B10;FALSE)
Here B1:B10 should be replaced with the cellrange of your new column that holds the true or false values
I have the array "A" with values:
101 101
0 0
61.6320000000000 0.725754779522671
73.7000000000000 0.830301150185882
78.2800000000000 0.490917508345341
81.2640000000000 0.602561200211232
82.6880000000000 0.435568593909153
And I wish to remove this first row and retain the shape of the array (2 columns), thus creating the array
0 0
61.6320000000000 0.725754779522671
73.7000000000000 0.830301150185882
78.2800000000000 0.490917508345341
81.2640000000000 0.602561200211232
82.6880000000000 0.435568593909153
I have used A = A(A~=101); , which removes the values as required - however it packs the array down to one column.
The best way is:
A = A(2:end, :)
But you can also do
A(1,:) = []
however it is slightly less efficient (see Deleting matrix elements by = [] vs reassigning matrix)
If you are looking to delete rows that equal a certain number try
A = A(A(:,1)~=101,:)
Use all or any if you want to delete row if either all or any column equals your value:
A = A(all(A~=101,2),:)
I've produced a code which separates data within a text file into the required format, filters the data and averages the output (in this case, the value in the fourth column)
I am trying to filter the data in column one for a list of values at the same time, with no strict pattern for the values. e.g 1001, 1007, 1048, 1192, 1200 ....
Currently my code only filters by a certain value (1001) is there a way of incorporating a list of values into this function?
C_f = C(C(:,1) == 1001 , :);
Any help would be much appreciated!
See if this is what you want,
val = [1000 1001];
ind = ismember(C(:,1),val);
C_f = C(ind,:)
What I have is a variable X which has values assigned to it in the form of a table of 9 columns and around 100 rows. Here is an example:
X =
Columns 1 through 7
-2.2869 -1.1168 0.1430 -4.0753 1.7620 -6.3229 -3.1997
-2.2504 -1.1022 0.2046 -3.9865 1.7423 -6.2172 -3.1231
-2.2138 -1.0876 0.2663 -3.8977 1.7226 -6.1115 -3.0465
-2.1772 -1.0730 0.3279 -3.8089 1.7029 -6.0058 -2.9700
I need to create a for loop that extracts the first r rows of the first 'p' colmuns. For example r=3 and p=4.
Any idea on how I can do that?
I suggest you don't use a for-loop, but rather index directly into the matrix:
out = X(1:r,1:p)
returns the first r rows and p columns of X.