一次重新缩放多个变量
Rescale multiple variable at once
我想一次重新调整多个变量。每个变量都应在 0 到 10 之间重新缩放。
我的数据集看起来像这样
df<-structure(list(Year = 1985:2012, r_mean_dp_C_EU_PTA = c(0.166685371371432,0, 0.340384674048008, 0.255663634111618, 0.137833312888481, 0.215940736735375,0.695926742038269, 1.12488458324014, 1.50426967770413, 1.96800275204271,
1.84220420613839, 2.55081439923073, 2.83958315572122, 3.02471358081631, 2.76227596053162, 5.13672466755955, 6.22501740311663, 6.04685020876299,
5.48990293535953, 5.74245144436088, 6.87554176822673, 5.35866756802216,6.21821261660873, 7.39740372167956, 7.37052059919359, 8.4053331043966,
7.88284279150424, 10),
r_mean_dp_C_US_PTA = c(0, 0.0243131684738152, 0.0295348762350131, 1.24572619158458, 1.20624633452509, 1.57418568231032,1.45479246796848, 2.38700784566208, 2.62865525326503, 2.26401361870534,2.67319203680329, 2.64440548764366, 3.10459526464658, 3.05231530072328,
3.32660416229216, 4.14909239351474, 3.76404440984403, 3.79766644256544,4.55279786294561, 5.57506946922008, 6.83412605593388, 8.07241989452914,9.10370786838265, 9.51564633960853, 8.64357423479438, 9.10723202296861,10, 9.06442082870898),
r_mean_dp_C_eu_esr_sum = c(0.0267071299038037,0, 0.0481033555876806, 0.039231355183461, 0.0255363040160583,0.0284158726695472, 0.234715155525714, 0.544954230234254, 0.683338138878583, 0.828929653572072, 0.950656658215744, 1.21492080702167, 1.30147631753441, 1.36122263965133, 1.33106989847101, 1.7848396827464, 2.19247065377408, 2.1506217173316, 4.91794342139369, 4.83398913690854, 7.28545175419305,5.42827409024432, 7.34375238832023, 8.91410171271897, 8.98533852868884, 9.17361943843028, 9.21421152468197, 10)), row.names = c(NA, -28L
),
class = c("data.table", "data.frame"))
我试过使用秤包，但没用
虽然带有名称标识符的函数失败
library(scales)
vars<-names(df[,2:4])
tst<-setDT(df)[, (vars):=lapply((vars), function(x) rescale(x,to = c(0,10)))]
使用位置标识符将所有变量值设置为 5，这不是我想要的。
tst<-setDT(df)[, 2:4:=lapply(2:4, function(x) rescale(x,to = c(0,10)))]
tst
# Year r_mean_dp_C_EU_PTA r_mean_dp_C_US_PTA r_mean_dp_C_eu_esr_sum
# 1: 1985 5 5 5
# 2: 1986 5 5 5
# 3: 1987 5 5 5
# 4: 1988 5 5 5
# 5: 1989 5 5 5
# 6: 1990 5 5 5
# 7: 1991 5 5 5
# 8: 1992 5 5 5
# 9: 1993 5 5 5
# 10: 1994 5 5 5
# 11: 1995 5 5 5
# 12: 1996 5 5 5
# 13: 1997 5 5 5
# 14: 1998 5 5 5
# 15: 1999 5 5 5
# 16: 2000 5 5 5
# 17: 2001 5 5 5
# 18: 2002 5 5 5
# 19: 2003 5 5 5
# 20: 2004 5 5 5
# 21: 2005 5 5 5
# 22: 2006 5 5 5
# 23: 2007 5 5 5
# 24: 2008 5 5 5
# 25: 2009 5 5 5
# 26: 2010 5 5 5
# 27: 2011 5 5 5
# 28: 2012 5 5 5
有人知道我做错了什么吗？
非常感谢您的帮助
我有点不清楚确切的输出需要是什么，因为在这个例子中一切都在 0 到 10 之间。
您是否尝试使用 dplyr？
tst <- df %>%
mutate_at(vars, function(x) rescale(x,to = c(0,10)) )
结果：
Year r_mean_dp_C_EU_PTA r_mean_dp_C_US_PTA r_mean_dp_C_eu_esr_sum
1 1985 0.1515322 0.00000000 0.02670713
2 1986 0.0000000 0.02431317 0.00000000
3 1987 0.3094406 0.02953488 0.04810336
4 1988 0.2324215 1.24572619 0.03923136
5 1989 0.1253030 1.20624633 0.02553630
6 1990 0.1963098 1.57418568 0.02841587
7 1991 0.6326607 1.45479247 0.23471516
8 1992 1.0226223 2.38700785 0.54495423
9 1993 1.3675179 2.62865525 0.68333814
10 1994 1.7890934 2.26401362 0.82892965
11 1995 1.6747311 2.67319204 0.95065666
12 1996 2.3189222 2.64440549 1.21492081
13 1997 2.5814392 3.10459526 1.30147632
14 1998 2.7497396 3.05231530 1.36122264
15 1999 2.5111600 3.32660416 1.33106990
16 2000 4.6697497 4.14909239 1.78483968
17 2001 5.6591067 3.76404441 2.19247065
18 2002 5.4971366 3.79766644 2.15062172
19 2003 4.9908209 4.55279786 4.91794342
20 2004 5.2204104 5.57506947 4.83398914
21 2005 6.2504925 6.83412606 7.28545175
22 2006 4.8715160 8.07241989 5.42827409
23 2007 5.6529206 9.10370787 7.34375239
24 2008 6.7249125 9.51564634 8.91410171
25 2009 6.7004733 8.64357423 8.98533853
26 2010 7.6412119 9.10723202 9.17361944
27 2011 7.1662207 10.00000000 9.21421152
28 2012 10.0000000 9.06442083 10.00000000
这是你想要的吗？
我们可以使用.SDcols。
按姓名申请
library(data.table)
df[, (vars):= lapply(.SD, scales::rescale, to = c(0, 10)), .SDcols = vars]
按职位申请
df[, 2:4 := lapply(.SD, scales::rescale, to = c(0, 10)), .SDcols = 2:4]
我想一次重新调整多个变量。每个变量都应在 0 到 10 之间重新缩放。 我的数据集看起来像这样
df<-structure(list(Year = 1985:2012, r_mean_dp_C_EU_PTA = c(0.166685371371432,0, 0.340384674048008, 0.255663634111618, 0.137833312888481, 0.215940736735375,0.695926742038269, 1.12488458324014, 1.50426967770413, 1.96800275204271,
1.84220420613839, 2.55081439923073, 2.83958315572122, 3.02471358081631, 2.76227596053162, 5.13672466755955, 6.22501740311663, 6.04685020876299,
5.48990293535953, 5.74245144436088, 6.87554176822673, 5.35866756802216,6.21821261660873, 7.39740372167956, 7.37052059919359, 8.4053331043966,
7.88284279150424, 10),
r_mean_dp_C_US_PTA = c(0, 0.0243131684738152, 0.0295348762350131, 1.24572619158458, 1.20624633452509, 1.57418568231032,1.45479246796848, 2.38700784566208, 2.62865525326503, 2.26401361870534,2.67319203680329, 2.64440548764366, 3.10459526464658, 3.05231530072328,
3.32660416229216, 4.14909239351474, 3.76404440984403, 3.79766644256544,4.55279786294561, 5.57506946922008, 6.83412605593388, 8.07241989452914,9.10370786838265, 9.51564633960853, 8.64357423479438, 9.10723202296861,10, 9.06442082870898),
r_mean_dp_C_eu_esr_sum = c(0.0267071299038037,0, 0.0481033555876806, 0.039231355183461, 0.0255363040160583,0.0284158726695472, 0.234715155525714, 0.544954230234254, 0.683338138878583, 0.828929653572072, 0.950656658215744, 1.21492080702167, 1.30147631753441, 1.36122263965133, 1.33106989847101, 1.7848396827464, 2.19247065377408, 2.1506217173316, 4.91794342139369, 4.83398913690854, 7.28545175419305,5.42827409024432, 7.34375238832023, 8.91410171271897, 8.98533852868884, 9.17361943843028, 9.21421152468197, 10)), row.names = c(NA, -28L
),
class = c("data.table", "data.frame"))
我试过使用秤包，但没用
虽然带有名称标识符的函数失败
library(scales)
vars<-names(df[,2:4])
tst<-setDT(df)[, (vars):=lapply((vars), function(x) rescale(x,to = c(0,10)))]
使用位置标识符将所有变量值设置为 5，这不是我想要的。
tst<-setDT(df)[, 2:4:=lapply(2:4, function(x) rescale(x,to = c(0,10)))]
tst
# Year r_mean_dp_C_EU_PTA r_mean_dp_C_US_PTA r_mean_dp_C_eu_esr_sum
# 1: 1985 5 5 5
# 2: 1986 5 5 5
# 3: 1987 5 5 5
# 4: 1988 5 5 5
# 5: 1989 5 5 5
# 6: 1990 5 5 5
# 7: 1991 5 5 5
# 8: 1992 5 5 5
# 9: 1993 5 5 5
# 10: 1994 5 5 5
# 11: 1995 5 5 5
# 12: 1996 5 5 5
# 13: 1997 5 5 5
# 14: 1998 5 5 5
# 15: 1999 5 5 5
# 16: 2000 5 5 5
# 17: 2001 5 5 5
# 18: 2002 5 5 5
# 19: 2003 5 5 5
# 20: 2004 5 5 5
# 21: 2005 5 5 5
# 22: 2006 5 5 5
# 23: 2007 5 5 5
# 24: 2008 5 5 5
# 25: 2009 5 5 5
# 26: 2010 5 5 5
# 27: 2011 5 5 5
# 28: 2012 5 5 5
有人知道我做错了什么吗？
非常感谢您的帮助
我有点不清楚确切的输出需要是什么，因为在这个例子中一切都在 0 到 10 之间。
您是否尝试使用 dplyr？
tst <- df %>%
mutate_at(vars, function(x) rescale(x,to = c(0,10)) )
结果：
Year r_mean_dp_C_EU_PTA r_mean_dp_C_US_PTA r_mean_dp_C_eu_esr_sum
1 1985 0.1515322 0.00000000 0.02670713
2 1986 0.0000000 0.02431317 0.00000000
3 1987 0.3094406 0.02953488 0.04810336
4 1988 0.2324215 1.24572619 0.03923136
5 1989 0.1253030 1.20624633 0.02553630
6 1990 0.1963098 1.57418568 0.02841587
7 1991 0.6326607 1.45479247 0.23471516
8 1992 1.0226223 2.38700785 0.54495423
9 1993 1.3675179 2.62865525 0.68333814
10 1994 1.7890934 2.26401362 0.82892965
11 1995 1.6747311 2.67319204 0.95065666
12 1996 2.3189222 2.64440549 1.21492081
13 1997 2.5814392 3.10459526 1.30147632
14 1998 2.7497396 3.05231530 1.36122264
15 1999 2.5111600 3.32660416 1.33106990
16 2000 4.6697497 4.14909239 1.78483968
17 2001 5.6591067 3.76404441 2.19247065
18 2002 5.4971366 3.79766644 2.15062172
19 2003 4.9908209 4.55279786 4.91794342
20 2004 5.2204104 5.57506947 4.83398914
21 2005 6.2504925 6.83412606 7.28545175
22 2006 4.8715160 8.07241989 5.42827409
23 2007 5.6529206 9.10370787 7.34375239
24 2008 6.7249125 9.51564634 8.91410171
25 2009 6.7004733 8.64357423 8.98533853
26 2010 7.6412119 9.10723202 9.17361944
27 2011 7.1662207 10.00000000 9.21421152
28 2012 10.0000000 9.06442083 10.00000000
这是你想要的吗？
我们可以使用.SDcols。
按姓名申请
library(data.table)
df[, (vars):= lapply(.SD, scales::rescale, to = c(0, 10)), .SDcols = vars]
按职位申请
df[, 2:4 := lapply(.SD, scales::rescale, to = c(0, 10)), .SDcols = 2:4]