while using Poco::format
I am trying to print the following line:
"3% and 5%"
int var1 = 3
int var2 = 5
std::string s;
s = format("%?d%% and %?d%%",var1,var2);
instead of getting "3% and 5%",
s equals "3 ?d%"
What am I doing wrong?
Related
I'm trying to divide the data by a certain datetime.
I've created e_timefrom what was originally a string "2019-10-15 20:33:04" for example.
To obtain all the information from the string containing h:m:s, I uses the following command to create a double
gen double e_time = clock(event_timestamp, "YMDhms")
Now I get the result I want from format e_time %tc (human readable),
I want to generate a new variable anything that is greater than 2019-10-15 as 1 and anything less than that as 0 .
I've tried
// 1
gen new_d = 0 if e_time < "1.887e+12"
replace new_d = 1 if e_time >= "1.887e+12"
// 2
gen new_d = 0 if e_time < "2019-10-15"
replace new_d = 1 if e_time > "2019-10-15"
However, I get an error message type mismatch.
I tried converting a string "2019-10-15" to double \to check if 1.887e+12 really meant 2019-10-15 using display, but I'm not sure how the command really works here.
Anyhow I tried
// 3
di clock("2019-10-15", "YMDhms")
but it didn't work.
Can anyone give advice on comparing dates that are in a double format properly?
Your post is a little hard to follow (a reproducible data example would help a lot) but the error type mismatch is because e_time is numeric, and "2019-10-15" is a string.
I suggest the following:
clear
input str20 datetime
"2019-10-14 20:33:04"
"2019-10-16 20:33:04"
end
* Keep first 10 characters
gen date = substr(datetime,1,10)
* Check that all strings are 10 characters
assert length(date) == 10
* Convert from string to numeric date variable
gen m = substr(date,6,2)
gen d = substr(date,9,2)
gen y = substr(date,1,4)
destring m d y, replace
gen newdate = mdy(m,d,y)
format newdate %d
gen wanted = newdate >= mdy(10,15,2019) & !missing(newdate)
drop date m d y
list
+------------------------------------------+
| datetime newdate wanted |
|------------------------------------------|
1. | 2019-10-14 20:33:04 14oct2019 0 |
2. | 2019-10-16 20:33:04 16oct2019 1 |
+------------------------------------------+
I am new to Nim, and wrote this simple code for fun:
var x: int = 3
var y: int = 4
if true:
y = 7
else:
x = 7
proc hello(xx: int, yy: int, ): int =
return xx + yy
hello(x, y)
The code seems fine (I checked with the Nim manuals), but it gives this weird error:
c:\Users\Xilpex\Desktop\Nim_tests\testrig.nim(12, 6) Error: expression 'hello(x, y)' is of type 'int' and has to be discarded
Why am I getting this error? Is there something I can do to fix it?
You are getting an error because procs declared to return values are meant to use that value somewhere, so the compiler reminds you that you are forgetting the result of the call. If some times you want the result, and others you want to ignore it, instead of creating a temporal variable you can use the discard statement or declare the proc as {.discardable.}.
I just found out why I was getting that error... It was because the procedure returned a value, and I wasn't storing that value anywhere. Here is the working code:
var x: int = 3
var y: int = 4
if true:
y = 7
else:
x = 7
proc hello(xx: int, yy: int, ): int =
return xx + yy
var output = hello(x, y)
I have a column of format DATETIME23. like this:
14.02.2017 13:00:25
I want to conver it to a string, so later, i would be able to modern it, so, for example, the final version would look like:
2017-02-14 13:00:25.000
Problem occures, when i try to convert date to char format: in result i have a string of smth like 1802700293 - which is the number of seconds.
I tried:
format date $23.0
or
date = put(date, $23.0)
P.S This is nother try:
data a;
format d date9.;
d = '12jan2016'd;
dtms = cat(day(d),'-',month(d),'-',year(d),' 00:00:00.000');
/* если нужно обязательно двухзначные день и месяц, то такой колхоз: */
if day(d) < 10 then dd=cat('0',put(day(d),$1.));
else ddday=put(day(d),$2.);
if month(d) < 10 then mm=cat('0',put(month(d),$1.));
else mm=put(month(d),$2.);
yyyy=put(year(d),$4.);
/*dtms2 = cat(dd,'-',mm,'-',yyyy,' 00:00:00.000');*/
dtms2 = cat(dd,'-',mm,'-',yyyy,' 00:00:00.000');
dtms = cat(day(d),'-',month(d),'-',year(d),' 00:00:00.000');
run;
BUT, abnormally, the dtms2 concat destroys the zero in the month element
If your datetime is stored as a SAS datetime, just use the appropriate format :
data test ;
dt = '09feb2017:13:53:26'dt ; /* specify a datetime constant */
new_dt = put(dt,E8601DT23.3) ; /* ISO datetime format */
run ;
Output
dt new_dt
1802267606 2017-02-09T13:53:26.000
If you need to replace the 'T' with a space, simply add a translate function around the put().
For your dtms solution you can use put and the Z2. format to keep the leading zero when you concatenate:
dtms = cat(day(d),'-', put(month(d),z2.),'-',year(d),' 00:00:00.000');
You should be able to just use put(date, datetime23.) for your problem though instead of $23, which is converting the number of seconds to a string with length 23. However, as a comment has mentioned datetime23. is not the format from your example.
I have a dataset with string variables and I am trying to generate a new binary variable based on the first two characters. All strings are 5 characters long, but I'm only concerned with the first two in order to sort.
For example, I could have 22001 and 22005. Since both are of the form 22XXX, I want to assign value 1 for both in the variable type_A. And if I have 25001 and 25005, since both are not of the form 22XXX, I want to assign value 0 for both in the variable type_A.
This should do the job:
clear
set obs 4
generate str5 var1 = "22001" in 1
replace var1 = "22005" in 2
replace var1 = "25001" in 3
replace var1 = "25005" in 4
gen type_A = substr(var1, 1, 2) == "22"
Please note that as you explain your problem it looks like you you are storing 22005 as text - which may not necessarily be the best idea..
For example I have: 1|2|3,4|5|6,7|8|10;
How can I output it like this:
A: 1 2 3
B: 4 5 6
C: 7 8 10
And how can I do this:
Array A = {1,2,3}
Array B = {4,5,6}
Array C = {7,8,10}
var reg = /(\d+)\|(\d+)\|(\d+)[,;]/g;
var str = "1|2|3,4|5|6,7|8|10;";
var index = 0;
str.replace(reg,function myfun(g,g1,g2,g3){
var ch = String.fromCharCode(65 + (index++));
return ch+": "+g1+" "+g2+" "+g3+"\n";
});
the second case should update myfun return string:
"Array "+ch+" = {"+g1+","+g2+","+g3+"}\n";