I am trying to calculate fsa variable values based on two discrete conditions. In the end, I want one output file with variable fsa. The first condition is fsa equals 1 when tas is greater than 275.15. In the second condition, I want fsa to be equal to (1-0.5*( tas -273.15)) when tas is between 273.15 and 275.15. Here is my first try using expr, any help is appreciated. any approach will be ok it doesn't have to be expr. my file size is big so a concise method is preferred.
cdo -expr, 'fsa = ((tas>275.15)) ? 1.0 : 0; fsa = ((tas<=275.15 & tas>273.15)) ? (1-0.5*(tas-273.15)) : fsa; <infile> <outfile>
Here is the answer by Ralf from CDO community main forum.
You can use nest the ternary operators like this:
cdo -expr,'fsa = ((tas<=275.15 && tas>273.15)) ? (1-0.5*(tas-273.15)) : ((tas>275.15) ? 1.0 : 0)' <infile> <outfile>
The logical OR is && !
Related
I'm trying to determine if it's between the hours of 12 am and 1 am. Here is my if statement:
If InStr(Time,"12") AND InStr(Time,"AM") Then
' Do something
Else
' Do something else
End If
The problem is that this statement evaluates to false, even if both of the conditions are true. I know this because I have tried a nested if like this
If InStr(Time,"12") Then
If InStr(Time,"AM") Then
' Do something
...
And that works. This also works
If InStr(Time,"12")<>0 AND InStr(Time,"AM")<>0 Then
' Do something
...
But if it works as a nested if, why can't I test both of the nested if conditions in a single if statement?
I replaced the InStr function calls with the values that they return
If 1 AND 10 Then
' Do something
Else
' Do something else
End If
And the same thing happened: the if statement evaluated as false and the "Do something else" commands were executed instead. But when I nested the second condition as another if statement inside the first if statement, the "Do something" commands were executed.
Why is that and is there any way to do this without the <>0 and without nesting?
If Time() >= TimeValue("12:00:00") AND Time() <= TimeValue("23:59:59") then
'Do Something
ElseIf Time() >= TimeValue("00:00:00") AND Time() <= TimeValue("01:00:00") then
'Do the same
Else
'Do something different
End If
This should work :)
The problem you observed is caused by the fact that VBScript uses the same operators for boolean and bit operations, depending on the data type of the operands. The InStr function returns a numeric value unless one of the strings is Null, so the operation becomes a bitwise comparison instead of a boolean comparison, as JosefZ pointed out. The behavior is documented:
The And operator also performs a bitwise comparison of identically positioned bits in two numeric expressions and sets the corresponding bit in result [...]
Demonstration:
>>> WScript.Echo "" & (True And True)
True
>>> WScript.Echo "" & (6 And 1) '0b0110 && 0b0001 ⇒ 0b0000
0
>>> WScript.Echo "" & (6 And 2) '0b0110 && 0b0010 ⇒ 0b0010
2
To enforce a boolean comparison you need to use InStr(...) > 0 or CBool(InStr(...)) (both of which evaluate to a boolean result) instead of just InStr(...) (which evaluates to a numeric result).
Date and Time are stored as number of days, where midnight is 0.0, and 1 am is 1/24 :
If Time <= 1/24 Then ' or If Time <= #1am# Then
When you using Time() function and if result like that 10:12:12 AM in this way Instr will result Ture because Instr by default use vbbinarycompare looking For any 12in binary format in 10:12:12 AM and there is sec and min 12 so it will Return True .
just try this :
myHour=replace(Time,Right(Time,9),"") 'get only the hour from time
myAMPM=replace(Time,Time,Right(Time,2)) 'get only AM or PM from time
If InStr(1,myHour,12,1) > 0 AND InStr(1,myAMPM,"AM",1) > 0 Then
wscript.echo "True"
Else
wscript.echo "False"
End If
I'm on Linux and I need to do an expr in order to match
6 digits with this range :
000001 to 999999
I'm stuck with '[0-9]{5}[1-9]' but I can't match numbers which end with 0 like 000010
I was thinking about '[0-9]{6}|?![0]{6}' in order to eliminate "000000"
How can I use ?! and/or are there any other solutions?
EDIT : solution = ((?!000000)[0-9]{6})
Using regex to check if a number is in a range isn't optimal. Instead, you can check for your inputs length and if it is in the range, using
a=000001
if ((${#a} == 6 && a > 0 && a <= 999999)); then
echo "foo"
fi
solution = ((?!000000)[0-9]{6})
I have a BigDecimal variable that contains number with minus(-) and not. I want to put static text that should be match with this conditions:
If $V{saldo} contains minus (-) value The static text will show "Rugi"
If $V {saldo} doesnt's contain minus(-) value the static text will show "Laba"
For example, if $V{saldo} = -250000 then the static text should be "Rugi", and if $V{saldo} = 400000 then the static text should be "Laba"
Try using this if-else construct:
$V{saldo}.intValue() < 0 ? "Rugi" : "Laba"
Please see this SO post for more information.
First Check for ZERO then apply your logic
$V{saldo}.$V{saldo} != 0.0 ? ($V{saldo}.intValue() < 0 ? "Rugi" : "Laba"): "Zero"
I have the following strings in a long string:
a=b=c=d;
a=b;
a=b=c=d=e=f;
I want to first search for above mentioned pattern (X=Y=...=Z) and then output like the following for each of the above mentioned strings:
a=d;
b=d;
c=d;
a=b;
a=f;
b=f;
c=f;
d=f;
e=f;
In general, I want all the variables to have an equal sign with the last variable on the extreme right of the string. Is there a way I can do it using regexprep in MATLAB. I am able to do it for a fixed length string, but for variable length, I have no idea how to achieve this. Any help is appreciated.
My attempt for the case of two equal signs is as follows:
funstr = regexprep(funstr, '([^;])+\s*=\s*+(\w+)+\s*=\s*([^;])+;', '$1 = $3; \n $2 = $3;\n');
Not a regexp but if you stick to Matlab you can make use of the cellfun function to avoid loop:
str = 'a=b=c=d=e=f;' ; %// input string
list = strsplit(str,'=') ;
strout = cellfun( #(a) [a,'=',list{end}] , list(1:end-1), 'uni', 0).' %'// Horchler simplification of the previous solution below
%// this does the same than above but more convoluted
%// strout = cellfun( #(a,b) cat(2,a,'=',b) , list(1:end-1) , repmat(list(end),1,length(list)-1) , 'uni',0 ).'
Will give you:
strout =
'a=f;'
'b=f;'
'c=f;'
'd=f;'
'e=f;'
Note: As Horchler rightly pointed out in comment, although the cellfun instruction allows to compact your code, it is just a disguised loop. Moreover, since it runs on cell, it is notoriously slow. You won't see the difference on such simple inputs, but keep this use when super performances are not a major concern.
Now if you like regex you must like black magic code. If all your strings are in a cell array from the start, there is a way to (over)abuse of the cellfun capabilities to obscure your code do it all in one line.
Consider:
strlist = {
'a=b=c=d;'
'a=b;'
'a=b=c=d=e=f;'
};
Then you can have all your substring with:
strout = cellfun( #(s)cellfun(#(a,b)cat(2,a,'=',b),s(1:end-1),repmat(s(end),1,length(s)-1),'uni',0).' , cellfun(#(s) strsplit(s,'=') , strlist , 'uni',0 ) ,'uni',0)
>> strout{:}
ans =
'a=d;'
'b=d;'
'c=d;'
ans =
'a=b;'
ans =
'a=f;'
'b=f;'
'c=f;'
'd=f;'
'e=f;'
This gives you a 3x1 cell array. One cell for each group of substring. If you want to concatenate them all then simply: strall = cat(2,strout{:});
I haven't had much experience w/ Matlab; but your problem can be solved by a simple string split function.
[parts, m] = strsplit( funstr, {' ', '='}, 'CollapseDelimiters', true )
Now, store the last part of parts; and iterate over parts until that:
len = length( parts )
for i = 1:len-1
print( strcat(parts(i), ' = ', parts(len)) )
end
I do not know what exactly is the print function in matlab. You can update that accordingly.
There isn't a single Regex that you can write that will cover all the cases. As posted on this answer:
https://stackoverflow.com/a/5019658/3393095
However, you have a few alternatives to achieve your final result:
You can get all the values in the line with regexp, pick the last value, then use a for loop iterating throughout the other values to generate the output. The regex to get the values would be this:
matchStr = regexp(str,'([^=;\s]*)','match')
If you want to use regexprep at any means, you should write a pattern generator and a replace expression generator, based on number of '=' in the input string, and pass these as parameters of your regexprep func.
You can forget about Regex and Split the input to generate the output looping throughout the values (similarly to alternative #1) .
I'm at a loss why the following code doesn't work. The intention is to input a vector of strings, some of which can be converted to a number, some can't. The following 'sapply' function should use a regex to match numbers and then return the number or (if not) return the original.
sapply(c("test","6","-99.99","test2"), function(v){
if(grepl("^[-+]?[0-9]*.?[0-9]+([eE][-+]?[0-9]+)?$",v)){as.numeric(v)} else {v}
})
Which returns the following result:
"test" "6" "-99.99" "test2"
Edit: What I expect the code to return:
"test" 6 -99.99 "test2
I can run the if statement on each element successfully.
> if(grepl("^[-+]?[0-9]*.?[0-9]+([eE][-+]?[0-9]+)?$","test")){as.numeric("test")} else {"test"}
[1] "test"
if(grepl("^[-+]?[0-9]*.?[0-9]+([eE][-+]?[0-9]+)?$","6")){as.numeric("6")} else {"6"}
[1] 6
And etc...
I don't understand why this is happening. I guess I have two questions. One: Why is this happening? And two: Usually I'm pretty good at troubleshooting, but I have no idea where to even look for this. If you know the problem, how did you find/know the solution? Should I open up the internal lapply function code?
that happens because sapply returns a vector, and a vector can't be mixed. If you use lapply then you get a list result which can be mixed, the same code but with lapply instead of sapply works how you want it to.
#Jeremy points into right direction, you can use lapply, which returns a list. Or, you can tell sapply not to simplify result.
If simplification occurs, the output type is determined from the
highest type of the return values in the hierarchy NULL < raw <
logical < integer < double < complex < character < list < expression,
after coercion of pairlists to lists.
out <- sapply(c("test","6","-99.99","test2"), function(v){
if(grepl("^[-+]?[0-9]*.?[0-9]+([eE][-+]?[0-9]+)?$",v)){
as.numeric(v)
} else {
v
}
}, simplify = FALSE)
> out
$test
[1] "test"
$`6`
[1] 6
$`-99.99`
[1] -99.99
$test2
[1] "test2"