I would like to round this figure to the nearest whole number. I am generating an xml based on an excel file, and would like to round the figure.
Here is my code:
xml.POS110 “wert”: “#{row[18]}”
I have tried:
xml.POS110 “wert”: “#{row[18]}”.round(0)
Move the round inside the quotes. Before, you were just trying to round a string. Also the default argument for round is 0 so you don't need to specify it (but you can if you really want to).
“#{row[18].round(0)}"
Related
I've managed to format the following lines in XPath, from this format:
1000.50
30
to this:
100050
3000
The solution I've adopted is:
concat(substring-before([number], '.'), substring-after([number], '.'))
If the . is not present I directly multiply the number by 100.
I'm wondering if there is any better way to do that. My second thought was using Java.
What goes wrong if you just multiply by 100? So long as the result of multiplying by 100 is an exact integer, it should be formatted without a "." when converted to a string. If there are rounding errors that mean the result is not an exact integer, you might want to use round().
The concat() approach seems fragile to me: what if someone gives you input like 1000.5 or perhaps 1000.500?
I'm trying to find a way to see if I can find a way to determine if a time that I stipulate falls between two other times. For example:
Start End
11:33:48 11:53:48
12:20:22 12:38:21
12:39:27 13:00:09
14:16:23 14:20:49
14:20:54 14:22:56
Then, I want to check if a cell (here the value of 12:50 in cell E30) falls between ANY two values in a range in THE SAME ROW. For me, I can get the obvious way to check for this in one row, and this simple version totally works:
=If(AND(E30>A4,E30<B4), "TRUE", "FALSE")
However, I want to check if that number falls within ANY of the values within the ROWS above cells in a range, and I can't get that to work. For example, I tried this and it didn't work:
=If(AND(E30>A:A,E30<B:B), "TRUE", "FALSE")
I also tried a simple countif variation but that didn't do it either:
=COUNTIFS(A:A,">"&E30,B:B,"<"&E30)
Any advice on how to adjust one of these formulas to get it to work?
Try switching the angle brackets around:
=COUNTIFS(A:A,"<"&E30,B:B,">"&E30)
I think this should work for the above data set -
=IF((FILTER(A2:B6, D2>A2:A6,D2<B2:B6)),TRUE,FALSE)
This will give you if there is any match or not.
For the number of rows count that match -
=ROWS((FILTER(A2:B6, D2>A2:A6,D2<B2:B6)))
Alomsot =IF(Q2>R2,IF(AND($X$16>HOUR(Q2),$X$16<(HOUR(R2)+12)),1,0),IF(AND(HOUR($X$14)>=HOUR(Q2),HOUR($X$14)<=HOUR(R2)),1,0))
I'm trying to figure out what is the most efficient way to parse data from a file using Lua. For example lets say I have a file (example.txt) with something like this in it:
0, Data
74, Instance
4294967295, User
255, Time
If I only want the numbers before the "," I could think of a few ways to get the information. I'd start out by getting the data with f = io.open(example.txt) and then use a for loop to parse each line of f. This leads to the heart of my question. What is the most efficient way to do this?
In the for loop I could use any of these methods to get the # before the comma:
line.find(regex)
line:gmatch(regex)
line:match(regex)
or Lua's split function
Has anyone run test for speed for these/other methods which they could point out as the fast way to parse? Bonus points if you can speak to speeds for parsing small vs. large files.
You probably want to use line:match("%d+").
line:find would work as well but returns more than you want.
line:gmatch is not what you need because it is meant to match several items in a string, not just one, and is meant to be used in a loop.
As for speed, you'll have to make your own measurements. Start with the simple code below:
for line in io.lines("example.txt") do
local x=line:match("%d+")
if x~=nil then print(x) end
end
I am attempting to use fortran to write out a comma-delimited file for import into another commercial package. The issue is that I have an unknown number of data columns. My output needs to look like this:
a_string,a_float,a_different_float,float_array_elem1,float_array_elem2,...,float_array_elemn
which would result in something that might look like this:
L1080,546876.23,4325678.21,300.2,150.125,...,0.125
L1090,563245.1,2356345.21,27.1245,...,0.00983
I have three issues. One, I would prefer the elements to be tightly grouped (variable column width), two, I do not know how to define a variable number of array elements in the format statement, and three, the array elements can span a large range--maybe 12 orders of magnitude. The following code conceptually does what I want, but the variable 'n' and the lack of column-width definition throws an error (of course):
WRITE(50,900) linenames(ii),loc(ii,1:2),recon(ii,1:n)
900 FORMAT(A,',',F,',',F,n(',',F))
(I should note that n is fixed at run-time.) The write statement does what I want it to when I do WRITE(50,*), except that it's width-delimited.
I think this thread almost answered my question, but I got quite confused: SO. Right now I have a shell script with awk fixing the issue, but that solution is...inelegant. I could do some manipulation to make the output a string, and then just write it, but I would rather like to avoid that option if at all possible.
I'm doing this in Fortran 90 but I like to try to keep my code as backwards-compatible as possible.
the format close to what you want is f0.3, this will give no spaces and a fixed number of decimal places. I think if you want to also lop off trailing zeros you'll need to do a good bit of work.
The 'n' in your write statement can be larger than the number of data values, so one (old school) approach is to put a big number there, eg 100000. Modern fortran does have some syntax to specify indefinite repeat, i'm sure someone will offer that up.
----edit
the unlimited repeat is as you might guess an asterisk..and is evideltly "brand new" in f2008
In order to make sure that no space occurs between the entries in your line, you can write them separately in character variables and then print them out using theadjustl() function in fortran:
program csv
implicit none
integer, parameter :: dp = kind(1.0d0)
integer, parameter :: nn = 3
real(dp), parameter :: floatarray(nn) = [ -1.0_dp, -2.0_dp, -3.0_dp ]
integer :: ii
character(30) :: buffer(nn+2), myformat
! Create format string with appropriate number of fields.
write(myformat, "(A,I0,A)") "(A,", nn + 2, "(',',A))"
! You should execute the following lines in a loop for every line you want to output
write(buffer(1), "(F20.2)") 1.0_dp ! a_float
write(buffer(2), "(F20.2)") 2.0_dp ! a_different_float
do ii = 1, nn
write(buffer(2+ii), "(F20.3)") floatarray(ii)
end do
write(*, myformat) "a_string", (trim(adjustl(buffer(ii))), ii = 1, nn + 2)
end program csv
The demonstration above is only for one output line, but you can easily write a loop around the appropriate block to execute it for all your output lines. Also, you can choose different numerical format for the different entries, if you wish.
I'm trying to read files and create a hashmap of the contents, but I'm having trouble at the parsing step. An example of the text file is
put 3
returns 3
between
3
pargraphs 1
4
3
#foo 18
****** 2
The word becomes the key and the number is the value. Notice that the spacing is fairly erratic. The word isn't always a word (which doesn't get picked up by /\w+/) and the number associated with that word isn't always on the same line. This is why I'm calling it not well-formed. If there were one word and one number on one line, I could just split it, but unfortunately, this isn't the case. I'm trying to create a hashmap like this.
{"put"=>3, "#foo"=>18, "returns"=>3, "paragraphs"=>1, "******"=>2, "4"=>3, "between"=>3}
Coming from Java, it's fairly easy. Using Scanner I could just use scanner.next() for the next key and scanner.nextInt() for the number associated with it. I'm not quite sure how to do this in Ruby when it seems I have to use regular expressions for everything.
I'd recommend just using split, as in:
h = Hash[*s.split]
where s is your text (eg s = open('filename').read. Believe it or not, this will give you precisely what you're after.
EDIT: I realized you wanted the values as integers. You can add that as follows:
h.each{|k,v| h[k] = v.to_i}