所以我希望能够基于df表找到ScoreLU值。例如,DSCRpd中的1.3730682的值应返回ScoreLU值60,因为它大于1.35但小于下一个值1.65。
另一方面,对于“杠杆”列,它必须按Desc顺序排列,即第一个值2.01应返回值60,因为它小于2.5但大于下一个值2.0。
[df][1]
DSCRpd Leverage TCB
1 1.3730682 2.010122 -1590099.11
2 1.0449597 2.680051 493370.85
3 1.0311141 4.790531 21594.63
4 1.3923007 3.279903 -499326.76
5 1.6443938 3.853003 988780.79
6 0.6265976 1.814359 1003736.73
7 2.1025253 4.412528 1245305.83
8 1.2872873 2.074424 -688305.83
9 0.5088294 2.504510 1406986.68
10 1.7794307 3.724905 1132513.33
[ScoreLU][2]
Score DSCRpd Leverage TCB
1: 0 0.65 5.0 0
2: 10 0.80 4.5 100000
3: 20 0.95 4.0 250000
4: 30 1.10 3.5 500000
5: 40 1.20 3.0 850000
6: 50 1.26 2.5 1250000
7: 60 1.35 2.0 1700000
8: 70 1.65 1.5 2300000
9: 80 2.00 1.0 2900000
10: 90 2.30 0.5 3600000
是的,就像具有Asc和Desc顺序功能的excel中的vlookup函数一样。帮助。
我有一个可以正确获取值的函数...但是我如何在每列上使用它来将值填充到适当的列,即,对于DSCRpd分数,结果应更新到名为DSCRpdScore的列。 / p>
此函数查看列号为CN的数据帧'df',并基于x返回适当的值。
myFUN = function(df, x, CN){
if (dtScoreLU[1,CN] <= median(dtScoreLU[,CN])){
myMax = max(dtScoreLU[(dtScoreLU[,CN] <= x),CN])
return(dtScoreLU %>% select(Score) %>%
filter(dtScoreLU[,CN] == myMax))
} else {
myMin = min(dtScoreLU[as.vector(dtScoreLU[,CN] >= x),CN])
return(dtScoreLU %>% select(Score) %>%
filter(dtScoreLU[,CN] == myMin))
}
}
答案 0 :(得分:0)
据我了解,这似乎是data.table
的滚动连接功能的不错选择。
所以我希望能够基于df表找到ScoreLU值。 例如,DSCRpd中的1.3730682的值应返回ScoreLU 值60,因为它大于1.35但小于下一个 值1.65。
library(data.table)
ScoreLU[, .(DSCRpd, Score)][df, ,on = 'DSCRpd', roll = TRUE]
DSCRpd Score Leverage TCB
1: 1.3730682 60 2.010122 -1590099.11
2: 1.0449597 20 2.680051 493370.85
3: 1.0311141 20 4.790531 21594.63
4: 1.3923007 60 3.279903 -499326.76
5: 1.6443938 60 3.853003 988780.79
6: 0.6265976 NA 1.814359 1003736.73
7: 2.1025253 80 4.412528 1245305.83
8: 1.2872873 50 2.074424 -688305.83
9: 0.5088294 NA 2.504510 1406986.68
10: 1.7794307 70 3.724905 1132513.33
另一方面,“杠杆”列需要按Desc顺序排列 即第一个值2.01应该按原样返回值60 小于2.5但大于下一个值2.0。
ScoreLU[, .(Leverage, Score)][df, , on = 'Leverage', roll = TRUE]
Leverage Score DSCRpd TCB
1: 2.010122 60 1.3730682 -1590099.11
2: 2.680051 50 1.0449597 493370.85
3: 4.790531 10 1.0311141 21594.63
4: 3.279903 40 1.3923007 -499326.76
5: 3.853003 30 1.6443938 988780.79
6: 1.814359 70 0.6265976 1003736.73
7: 4.412528 20 2.1025253 1245305.83
8: 2.074424 60 1.2872873 -688305.83
9: 2.504510 50 0.5088294 1406986.68
10: 3.724905 30 1.7794307 1132513.33
如果愿意,可以将它们组合在一起
ScoreLU[, .(Leverage, Score)][
ScoreLU[, .(DSCRpd, Score)][
df, ,on = 'DSCRpd', roll = TRUE
], , on = 'Leverage', roll = TRUE]
Leverage Score DSCRpd i.Score TCB
1: 2.010122 60 1.3730682 60 -1590099.11
2: 2.680051 50 1.0449597 20 493370.85
3: 4.790531 10 1.0311141 20 21594.63
4: 3.279903 40 1.3923007 60 -499326.76
5: 3.853003 30 1.6443938 60 988780.79
6: 1.814359 70 0.6265976 NA 1003736.73
7: 4.412528 20 2.1025253 80 1245305.83
8: 2.074424 60 1.2872873 50 -688305.83
9: 2.504510 50 0.5088294 NA 1406986.68
10: 3.724905 30 1.7794307 70 1132513.33
对于两个Score
变量的末尾,可以根据需要指定rollends
自变量。如果您有时间,我也会给?data.table
做通读。这对入门很有帮助,因为语法有时可能有点不透明。
我对data.table
还是陌生的,所以欢迎其他有更多专业知识的人加入。
ScoreLU <- structure(list(Score = c(0L, 10L, 20L, 30L, 40L, 50L, 60L, 70L, 80L, 90L),
DSCRpd = c(0.65, 0.8, 0.95, 1.1, 1.2, 1.26, 1.35, 1.65, 2, 2.3),
Leverage = c(5, 4.5, 4, 3.5, 3, 2.5, 2, 1.5, 1, 0.5),
TCB = c(0L, 100000L, 250000L, 500000L, 850000L, 1250000L, 1700000L, 2300000L, 2900000L, 3600000L)),
.Names = c("Score", "DSCRpd", "Leverage", "TCB"), row.names = c(NA, -10L), class = c("data.table", "data.frame"))
df <- structure(list(DSCRpd = c(1.3730682, 1.0449597, 1.0311141, 1.3923007, 1.6443938, 0.6265976, 2.1025253, 1.2872873, 0.5088294, 1.7794307),
Leverage = c(2.010122, 2.680051, 4.790531, 3.279903, 3.853003, 1.814359, 4.412528, 2.074424, 2.50451, 3.724905),
TCB = c(-1590099.11, 493370.85, 21594.63, -499326.76, 988780.79, 1003736.73, 1245305.83, -688305.83, 1406986.68, 1132513.33)),
.Names = c("DSCRpd", "Leverage", "TCB"),
row.names = c(NA, -10L), class = c("data.table", "data.frame" ))