library(data.table)
set.seed(01)
DATA = data.table("STUDENT" = c(1:2000),
"GROUP" = c(sample(letters[1:3], size=2000, r = T)),
"CLASS" = c(sample(1:6, size=2000, r = T)),
"SCORE1" = c(sample(50:100,size=2000,r=T)),
"SCORE2" = c(sample(40:100, size=2000, r = T)))
Say you have this data and you wish to make SEVERAL extra data.table that are aggregated or collapse...
a) mean SCORE1 and mean SCORE2 by GROUP
b) mean SCORE1 and mean SCORE2 by CLASS
c) mean SCORE1 and mean SCORE2 by GROUP and by CLASS
d) count of GROUP by CLASS
Is there more efficient way to use data.table to create these 4 data.table? Or do you simply recommend making all of them and storing as 4 separate data.tables?
CodePudding user response：
They all have different structures
(a)
DATA[, .(mSCORE1 = mean(SCORE1), mSCORE2= mean(SCORE2)), by=.(GROUP)]
GROUP mSCORE1 mSCORE2
1: a 74.59971 69.98240
2: c 75.12943 71.30817
3: b 74.95964 69.97608
(b and d combined)
DATA[, .(mSCORE1 = mean(SCORE1), mSCORE2= mean(SCORE2),nGROUPS =uniqueN(GROUP)), by=.(CLASS)]
CLASS mSCORE1 mSCORE2 nGROUPS
1: 6 75.01685 70.14607 3
2: 3 75.76025 70.18927 3
3: 4 74.25532 70.74164 3
4: 5 75.77126 69.46921 3
5: 1 73.28797 71.56646 3
6: 2 75.17595 70.44282 3
(c)
DATA[, lapply(.SD, mean, na.rm=T), by=.(GROUP,CLASS),.SDcols = patterns("SC")]
GROUP CLASS SCORE1 SCORE2
1: a 6 75.27027 69.00901
2: c 3 78.05660 70.18868
3: a 4 75.72727 68.95868
4: b 6 74.20455 70.78788
5: a 5 75.94915 69.78814
6: c 1 73.93043 72.63478
7: c 2 75.18812 71.80198
8: b 2 76.46667 67.89167
9: c 5 75.28814 70.63559
10: a 1 72.43519 69.72222
11: a 2 73.87500 71.85000
12: b 1 73.48387 72.38710
13: b 5 76.11429 67.80000
14: b 3 75.07477 69.84112
15: c 6 75.71681 70.51327
16: b 4 74.26786 71.41964
17: a 3 74.12500 70.54808
18: c 4 72.38542 72.19792
If you want to get all these in one table, of course you would have class-specific summaries and group-specific summaries replicated across rows of group and class. Here is an illustration in a single pipeline
DATA[, .(mSCORE1_grpclass = mean(SCORE1), mSCORE2_grpclass= mean(SCORE2), NSTUDENTS=.N), by=.(GROUP,CLASS)] %>%
.[,`:=`(
mSCORE1_grp=sum(NSTUDENTS*mSCORE1_grpclass)/sum(NSTUDENTS),
mSCORE2_grp=sum(NSTUDENTS*mSCORE2_grpclass)/sum(NSTUDENTS)),by=.(GROUP)] %>%
.[,`:=`(
mSCORE1_class=sum(NSTUDENTS*mSCORE1_grpclass)/sum(NSTUDENTS),
mSCORE2_class=sum(NSTUDENTS*mSCORE2_grpclass)/sum(NSTUDENTS),
nGROUPS_class = uniqueN(GROUP)),by=.(CLASS)] %>%
.[]
Output:
GROUP CLASS mSCORE1_grpclass mSCORE2_grpclass NSTUDENTS mSCORE1_grp mSCORE2_grp mSCORE1_class mSCORE2_class nGROUPS_class
1: a 6 75.27027 69.00901 111 74.59971 69.98240 75.01685 70.14607 3
2: c 3 78.05660 70.18868 106 75.12943 71.30817 75.76025 70.18927 3
3: a 4 75.72727 68.95868 121 74.59971 69.98240 74.25532 70.74164 3
4: b 6 74.20455 70.78788 132 74.95964 69.97608 75.01685 70.14607 3
5: a 5 75.94915 69.78814 118 74.59971 69.98240 75.77126 69.46921 3
6: c 1 73.93043 72.63478 115 75.12943 71.30817 73.28797 71.56646 3
7: c 2 75.18812 71.80198 101 75.12943 71.30817 75.17595 70.44282 3
8: b 2 76.46667 67.89167 120 74.95964 69.97608 75.17595 70.44282 3
9: c 5 75.28814 70.63559 118 75.12943 71.30817 75.77126 69.46921 3
10: a 1 72.43519 69.72222 108 74.59971 69.98240 73.28797 71.56646 3
11: a 2 73.87500 71.85000 120 74.59971 69.98240 75.17595 70.44282 3
12: b 1 73.48387 72.38710 93 74.95964 69.97608 73.28797 71.56646 3
13: b 5 76.11429 67.80000 105 74.95964 69.97608 75.77126 69.46921 3
14: b 3 75.07477 69.84112 107 74.95964 69.97608 75.76025 70.18927 3
15: c 6 75.71681 70.51327 113 75.12943 71.30817 75.01685 70.14607 3
16: b 4 74.26786 71.41964 112 74.95964 69.97608 74.25532 70.74164 3
17: a 3 74.12500 70.54808 104 74.59971 69.98240 75.76025 70.18927 3
18: c 4 72.38542 72.19792 96 75.12943 71.30817 74.25532 70.74164 3