Hi all,
I have been chipping away at a problem I encountered in calculating
rates per year from a moderately large data file (46412 rows). When I
ran the following command, I got obviously wrong output:
interval<-
c(NA,as.numeric(diff(
strptime(mkdf$MEAS_DATE,"%d/%m/%Y")))/365.25)
The values in MEAS_DATE looked like this:
mkdf$MEAS_DATE[1:10]
[1] 1/5/1962 1/5/1963 1/5/1964 1/3/1965 1/4/1966 1/4/1967
1/6/1968
[8] 25/3/1969 1/4/1971 1/2/1974
146 Levels: 10/10/1967 1/10/1947 1/10/1965 1/10/1967 1/10/1983 ... 9/1/1992
To abbreviate three evenings of work, I finally found that values 17170
and 17171 were the same. If I ran the entire set, or anything over
1:17170, I would get output like this:
interval[1:10]
[1] NA 86340.86 86577.41 71911.29 93673.92 86340.86
101006.98
[8] 70255.44 174337.58 245292.81
If I ran any set of values up to 17170, I would get the correct output:
interval[1:10]
[1] NA 0.9993155 1.0020534 0.8323066 1.0841889 0.9993155
1.1690623
[8] 0.8131417 2.0177960 2.8390372
If I changed value 17171 by one day (and added that level), the command
worked correctly:
interval[1:10]
[1] NA 0.9993155 1.0020534 0.8323066 1.0841889 0.9993155
1.1690623
[8] 0.8131417 2.0177960 2.8390372
There have been a few messages about this problem, but apparently no
solution. The problem can be seen with these examples (I haven't
included the real data as it is not mine):
foodate<-c("1/7/1991","1/8/1991","1/8/1991","3/8/1991")
as.numeric(diff(strptime(foodate,"%d/%m/%Y"))/365.25)
[1] 7333.0595 0.0000 473.1006
foodate<-factor(c("1/7/1991","1/8/1991","1/8/1991","3/8/1991"))
as.numeric(diff(strptime(foodate,"%d/%m/%Y"))/365.25)
[1] 7333.0595 0.0000 473.1006
foodate<-factor(c("1/7/1991","1/8/1991","2/8/1991","3/8/1991"))
> as.numeric(diff(strptime(foodate,"%d/%m/%Y"))/365.25)
[1] 0.084873374 0.002737851 0.002737851
Beats me.
Jim
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