数据集：
zip acs.pop napps pperct cgrp zgrp perc
1: 12007 97 2 2.0618557 2 1 25.000000
2: 12007 97 2 2.0618557 NA 2 50.000000
3: 12007 97 2 2.0618557 1 1 25.000000
4: 12008 485 2 0.4123711 2 1 33.333333
5: 12008 485 2 0.4123711 4 1 33.333333
6: 12008 485 2 0.4123711 NA 1 33.333333
7: 12009 7327 187 2.5522042 4 76 26.206897
8: 12009 7327 187 2.5522042 1 41 14.137931
9: 12009 7327 187 2.5522042 2 23 7.931034
10: 12009 7327 187 2.5522042 NA 103 35.517241
11: 12009 7327 187 2.5522042 3 47 16.206897
12: 12010 28802 580 2.0137490 NA 275 32.163743
13: 12010 28802 580 2.0137490 4 122 14.269006
14: 12010 28802 580 2.0137490 1 269 31.461988
15: 12010 28802 580 2.0137490 2 96 11.228070
16: 12010 28802 580 2.0137490 3 93 10.877193
17: 12018 7608 126 1.6561514 3 30 16.129032
18: 12018 7608 126 1.6561514 NA 60 32.258065
19: 12018 7608 126 1.6561514 2 14 7.526882
20: 12018 7608 126 1.6561514 4 57 30.645161
21: 12018 7608 126 1.6561514 1 25 13.440860
22: 12019 14841 144 0.9702850 NA 62 30.097087
23: 12019 14841 144 0.9702850 4 73 35.436893
24: 12019 14841 144 0.9702850 3 30 14.563107
25: 12019 14841 144 0.9702850 1 23 11.165049
26: 12019 14841 144 0.9702850 2 18 8.737864
27: 12020 31403 343 1.0922523 3 76 14.960630
28: 12020 31403 343 1.0922523 1 88 17.322835
29: 12020 31403 343 1.0922523 2 38 7.480315
30: 12020 31403 343 1.0922523 4 141 27.755906
31: 12020 31403 343 1.0922523 NA 165 32.480315
32: 12022 1002 5 0.4990020 NA 4 44.444444
33: 12022 1002 5 0.4990020 4 2 22.222222
34: 12022 1002 5 0.4990020 3 1 11.111111
35: 12022 1002 5 0.4990020 1 1 11.111111
我知道reshape2或reshape包可以处理这个问题，但我不确定如何处理。我需要最终输出看起来像这样：
zip acs.pop napps pperct zgrp4 zgrp3 zgrp2 zgrp1 perc4 perc3 perc2 perc1
12009 7327 187 2.5522042 76 47 23 41 26.206897 16.206897 7.931034 14.137931
zip是id
acs.pop，napps，pperct对于每个zip组都是相同的
zgrp4 ... zgrp1是cgrp
每个值的zgrp值
perc4 ... perc1是cgrp
答案 0 :(得分：5)
我们可以从dcast的devel版本尝试data.table，它可以包含多个value.var列。在这种情况下，我们有＆＃39; zgrp＆＃39;和＆＃39; perc＆＃39;是值列。使用分组变量，我们创建一个序列变量（＆＃39; ind＆＃39;），然后使用dcast转换来自＆＃39; long＆＃39;广泛的＆＃39;格式。
安装devel版本的说明是here
library(data.table)#v1.9.5
setDT(df1)[, ind:= 1:.N, .(zip, acs.pop, napps, pperct)]
dcast(df1, zip+acs.pop + napps+pperct~ind, value.var=c('zgrp', 'perc'))
# zip acs.pop napps pperct 1_zgrp 2_zgrp 3_zgrp 4_zgrp 5_zgrp 1_perc
#1: 12007 97 2 2.0618557 1 2 1 NA NA 25.00000
#2: 12008 485 2 0.4123711 1 1 1 NA NA 33.33333
#3: 12009 7327 187 2.5522042 76 41 23 103 47 26.20690
#4: 12010 28802 580 2.0137490 275 122 269 96 93 32.16374
#5: 12018 7608 126 1.6561514 30 60 14 57 25 16.12903
#6: 12019 14841 144 0.9702850 62 73 30 23 18 30.09709
#7: 12020 31403 343 1.0922523 76 88 38 141 165 14.96063
#8: 12022 1002 5 0.4990020 4 2 1 1 NA 44.44444
# 2_perc 3_perc 4_perc 5_perc
#1: 50.00000 25.000000 NA NA
#2: 33.33333 33.333333 NA NA
#3: 14.13793 7.931034 35.51724 16.206897
#4: 14.26901 31.461988 11.22807 10.877193
#5: 32.25807 7.526882 30.64516 13.440860
#6: 35.43689 14.563107 11.16505 8.737864
#7: 17.32284 7.480315 27.75591 32.480315
#8: 22.22222 11.111111 11.11111 NA
或者我们可以使用ave/reshape
base R
df2 <- transform(df1, ind=ave(seq_along(zip), zip,
acs.pop, napps, pperct, FUN=seq_along))
reshape(df2, idvar=c('zip', 'acs.pop', 'napps', 'pperct'),
timevar='ind', direction='wide')
答案 1 :(得分：1)
这是spread()中tidyr的好用途。
df %>% filter(!is.na(cgrp)) %>% # if cgrp is missing I don't know where to put the obs
gather(Var, Val,6:7) %>% # one row per measure (zgrp OR perc) observed
group_by(zip, acs.pop, napps, pperct) %>% # unique combos of these will define rows in output
unite(Var1,Var,cgrp) %>% # indentify which obs for which measure
spread(Var1, Val) # make columns for zgrp_1, _2, etc., perc1,2, etc
示例输出：
> df2[df2$zip==12009,]
Source: local data frame [1 x 12]
zip acs.pop napps pperct perc_1 perc_2 perc_3 perc_4 zgrp_1 zgrp_2 zgrp_3 zgrp_4
1 12009 7327 187 2.552204 14.13793 7.931034 16.2069 26.2069 41 23 47 76
感谢@akrun的帮助