Displaying 3 results from an estimated 3 matches for "20080505l".
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20080505
2009 Mar 31
0
data.table is on CRAN (enhanced data.frame for time series joins and more)
...=3),
date=c(20080501L,20080502L,20080506L), v=1:6)
setkey(DT,id,date)
DT["A"] # all 3 rows for A since mult
by default is "all"
DT[J("A",20080502L)] # row for A where date also matches
exactly
DT[J("A",20080505L)] # NA since 5 May is missing (outer join
by default)
DT[J("A",20080505L),nomatch=0] # inner join instead
dts = c(20080501L, 20080502L, 20080505L, 20080506L, 20080507L, 20080508L)
DT[J("A",dts)] # 3 of the dates in dts mat...
2009 Mar 31
0
data.table is on CRAN (enhanced data.frame for time series joins and more)
...=3),
date=c(20080501L,20080502L,20080506L), v=1:6)
setkey(DT,id,date)
DT["A"] # all 3 rows for A since mult
by default is "all"
DT[J("A",20080502L)] # row for A where date also matches
exactly
DT[J("A",20080505L)] # NA since 5 May is missing (outer join
by default)
DT[J("A",20080505L),nomatch=0] # inner join instead
dts = c(20080501L, 20080502L, 20080505L, 20080506L, 20080507L, 20080508L)
DT[J("A",dts)] # 3 of the dates in dts mat...
2009 Apr 01
0
回复： R-help Digest, Vol 73, Issue 32
...=3),
date=c(20080501L,20080502L,20080506L), v=1:6)
setkey(DT,id,date)
DT["A"] # all 3 rows for A since mult
by default is "all"
DT[J("A",20080502L)] # row for A where date also matches
exactly
DT[J("A",20080505L)] # NA since 5 May is missing (outer join
by default)
DT[J("A",20080505L),nomatch=0] # inner join instead
dts = c(20080501L, 20080502L, 20080505L, 20080506L, 20080507L, 20080508L)
DT[J("A",dts)] # 3 of the dates in dts mat...