根据唯一值和列值从数据框中随机绘制行 [英] Randomly draw rows from dataframe based on unique values and column values

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本文介绍了根据唯一值和列值从数据框中随机绘制行的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有一个包含许多描述符变量(trt,个人,会话)的数据框。我希望能够随机选择可能的 trt x个人组合的一部分,但要控制会话变量,以使随机抽取的会话数不相同。这是我的数据帧的样子:

  trt<-c(rep(c(rep( A,3) ,rep( B,3),rep( C,3)),9))
个人<-rep(c( Bob, Nancy, Tim),27)
会话<-rep(1:27,每个= 3)
数据<-rnorm(81,平均值= 4,sd = 1)
df<-data.frame( trt,个人,会话,数据))
df
trt个人会话数据
1 A Bob 1 3.72013685581385
2 A Nancy 1 3.97225419000673
3 A Tim 1 4.44714175686225
4 B Bob 2 5.00024599458127
5 B Nancy 2 3.43615965145765
6 B Tim 2 6.7920094635501
7 C Bob 3 4.36315054477571
8 C Nancy 3 5.07117348146375
9 C Tim 3 4.38503325758969
10 A Bob 4 4.30677162933005
11 A Nancy 4 1.89311687510669
12 A Tim 4 3.09084920968413
13 B Bob 5 3.10436190897 144
14 B南希5 3.59454992439722
15 B蒂姆5 3.40778069131207
16 C鲍勃6 4.00171937800892
17辰南希6 0.14578811080644
18蒂姆6 4.20754733296227
19 A Bob 7 3.69131009783284
20 A Nancy 7 4.7025756891679
21 A Tim 7 4.46196017363017
22 B Bob 8 3.97573281432736
23 B Nancy 8 4.5373185942686
24 B Tim 8 2.40937847038141
25 C Bob 9 4.57519884980087
26 C Nancy 9 5.19143914630448
27 C Tim 9 4.83144732833874
28 A Bob 10 3.01769965527235
29 A Nancy 10 5.17300616827746
30 A Tim 10 4.65432284571663
31 B Bob 11 4.50892032922527
32 B Nancy 11 3.38082717995663
33 B Tim 11 4.92022245677209
34 C Bob 12 4.541 49796547394
35 C Nancy 12 3.21992774137179
36 C Tim 12 3.74507360931023
37 A Bob 13 3.39524949548056
38 A Nancy 13 4.17518916890901
39 A Tim 13 3.02932375225388
40 B Bob 14 3.59660910672907
41 B Nancy 14 2.08784850191654
42 B Tim 14 3.98446125755258
43 C Bob 15 4.01837496797085
44 C Nancy 15 3.40610126858125
45 C Tim 15 4.57107635588582
46 A Bob 16 3.15839276840723
47 A Nancy 16 2.19932140340504
48 A Tim 16 4.77588798035668
49 B Bob 17 4.3524768657397
50 B Nancy 17 4.49071625925856
51 B Tim 17 4.02576463486266
52 C Bob 18 3.74783360762117
53 C Nancy 18 2.84123227236184
54 C Tim 18 3.2024114782253
55 A鲍勃19 4.93837445490921
56 A Nancy 19 4.7103051496802
57 A Tim 19 6.22083635045134
58 B Bob 20 4.5177747677824
59 B Nancy 20 1.78839270771153
60 B Tim 20 5.07140678136995
61 C Bob 21 3.47818616035335
62 C Nancy 21 4.28526474048439
63 C Tim 21 4.22597602946575
64 A Bob 22 1.91700925257901
65 A Nancy 22 2.96317997587458
66 A Tim 22 2.53506974227672
67 B Bob 23 5.52714403395316
68 B Nancy 23 3.3618513551059
69 B Tim 23 4.85869007113978
70 C Bob 24 3.4367068543959
71 C Nancy 24 4.47769879000349
72 C Tim 24 5.77340483757836
73 A Bob 25 4.78524317734622
74 A Nancy 25 3.55373702554664
75 A Tim 25 2.88541465503637
76 BB ob 26 4.62885302019139
77 B南希26 3.59430293369092
78 B Tim 26 2.29610255924296
79 C Bob 27 4.38433001299722
80 C Nancy 27 3.77825207859976
81 C Tim 27 2.12163194694365

如何从每个 trt x个人中抽取2个具有唯一会话号的组合?这是一个示例,我希望数据帧看起来像这样:

  trt个人会话数据
1 A Bob 1 3.72013685581385
5 B Nancy 2 3.43615965145765
7 C Bob 3 4.36315054477571
12 A Tim 4 3.09084920968413
15 B Tim 5 3.40778069131207
17 C Nancy 6 0.14578811080644
19 A Bob 7 3.69131009783284
29 A Nancy 10 5.17300616827746
31 B Bob 11 4.50892032922527
34 C Bob 12 4.54149796547394
39 A Tim 13 3.02932375225388
40 B Bob 14 3.59660910672907
47 A Nancy 16 2.19932140340504
51 B Tim 17 4.02576463486266
54 C Tim 18 3.2024114782253
59 B Nancy 20 1.78839270771153
71 C Nancy 24 4.47769879000349
81 C Tim 27 2.12163194694365

我尝试了几件事没有运气。 / p>

我试图随机选择两个 trt x单个组合,但最终得到重复的会话值:

  setDT((df))
df [,.SD [sample(.N,2)],keyby = 。((trt,个人)]
trt个人会话数据
1:A Bob 25 2.7560788894668
2:A Bob 19 4.12040841647523
3:A Nancy 4 5.35362338127901
4 :A Nancy 19 5.51636882737692
5:A Tim 19 5.10553640201998
6:A Tim 1 2.77380671625473
7:B Bob 23 3.50585105164409
8:B Bob 8 3.58167259470814
9 :B南希23 2.85301307507985
10:B南希8 2.85179395539781
11:B蒂姆26 2.40666507132474
12:B Tim 20 3.31276311351286
13:C Bob 24 3.19076007024549
14:C Bob 3 3.59146613276121
15:C Nancy 9 4.46606667880457
16:C Nancy 15 2.25405252536256
17:C Tim 12 4.43111661206133
18:C Tim 27 4.23868848646589

我尝试随机选择每个会话号,然后提取2个 trt x个人组合,但是由于随机选择没有抓住相等数量的 trt x,通常会返回错误单个组合:

  ind<-sapply(unique(df $ session),function(x )sample(which(df $ session == x),1))
df.unique<-df [ind,]
df.sub<-df.unique [,.SD [sample (.N,2)],按=。(trt,单个)]
`[.data.frame`(df.unique,,.SD [sample(.N,2)]]中的错误,由= 。((trt,个)):
未使用的参数(by =。(trt,个))

预先感谢您的帮助!

解决方案

也许是一种聪明的采样方式,但同时有一个简单的主意:

  setDT(df)
setkey(df,session)

usedsessions = 0#一些不是会话号的值
df [,{
res = .SD [!。(usedsessions)] [sample(.N,2)]
usedsessions = c(已使用会话,res $ session)
res
}
,由=。(trt,单个)]
#trt个人会话数据
#1:一个Bob 7 4.256668
#2:一个Bob 25 2.431821
#3:一个Nancy 16 4.785859
#4:一个Nancy 19 4.865248
#5:A Tim 4 3.303689
#6:A Tim 13 3.550261
#7:B Bob 26 3.987136
#8:B Bob 17 3.283055
#9 :B南希14 3.177226
#10:B南希2 3.639542
#11: B Tim 8 2.168447
#12:B Tim 5 3.521123
#13:C Bob 21 3.284245
#14:C Bob 12 5.773098
#15:C Nancy 24 4.624428
#16:C Nancy 9 3.235467
#17:C Tim 18 4.001395
#18:C Tim 27 5.002110

您可能需要添加特殊情况处理(例如如果没有这样的抽样)。


I have a dataframe with many descriptor variables (trt, individual, session). I want to be able to randomly select a fraction of the possible trt x individual combinations but control for the session variable such that no random pull has the same session number. Here is what my dataframe looks like:

trt <- c(rep(c(rep("A", 3), rep("B", 3), rep("C", 3)), 9))
individual <- rep(c("Bob", "Nancy", "Tim"), 27)
session <- rep(1:27, each = 3)
data <- rnorm(81, mean = 4, sd = 1)
df <- data.frame(trt, individual, session, data))
df
   trt individual session             data
1    A        Bob       1 3.72013685581385
2    A      Nancy       1 3.97225419000673
3    A        Tim       1 4.44714175686225
4    B        Bob       2 5.00024599458127
5    B      Nancy       2 3.43615965145765
6    B        Tim       2  6.7920094635501
7    C        Bob       3 4.36315054477571
8    C      Nancy       3 5.07117348146375
9    C        Tim       3 4.38503325758969
10   A        Bob       4 4.30677162933005
11   A      Nancy       4 1.89311687510669
12   A        Tim       4 3.09084920968413
13   B        Bob       5 3.10436190897144
14   B      Nancy       5 3.59454992439722
15   B        Tim       5 3.40778069131207
16   C        Bob       6 4.00171937800892
17   C      Nancy       6 0.14578811080644
18   C        Tim       6 4.20754733296227
19   A        Bob       7 3.69131009783284
20   A      Nancy       7  4.7025756891679
21   A        Tim       7 4.46196017363017
22   B        Bob       8 3.97573281432736
23   B      Nancy       8  4.5373185942686
24   B        Tim       8 2.40937847038141
25   C        Bob       9 4.57519884980087
26   C      Nancy       9 5.19143914630448
27   C        Tim       9 4.83144732833874
28   A        Bob      10 3.01769965527235
29   A      Nancy      10 5.17300616827746
30   A        Tim      10 4.65432284571663
31   B        Bob      11 4.50892032922527
32   B      Nancy      11 3.38082717995663
33   B        Tim      11 4.92022245677209
34   C        Bob      12 4.54149796547394
35   C      Nancy      12 3.21992774137179
36   C        Tim      12 3.74507360931023
37   A        Bob      13 3.39524949548056
38   A      Nancy      13 4.17518916890901
39   A        Tim      13 3.02932375225388
40   B        Bob      14 3.59660910672907
41   B      Nancy      14 2.08784850191654
42   B        Tim      14 3.98446125755258
43   C        Bob      15 4.01837496797085
44   C      Nancy      15 3.40610126858125
45   C        Tim      15 4.57107635588582
46   A        Bob      16 3.15839276840723
47   A      Nancy      16 2.19932140340504
48   A        Tim      16 4.77588798035668
49   B        Bob      17  4.3524768657397
50   B      Nancy      17 4.49071625925856
51   B        Tim      17 4.02576463486266
52   C        Bob      18 3.74783360762117
53   C      Nancy      18 2.84123227236184
54   C        Tim      18  3.2024114782253
55   A        Bob      19 4.93837445490921
56   A      Nancy      19  4.7103051496802
57   A        Tim      19 6.22083635045134
58   B        Bob      20  4.5177747677824
59   B      Nancy      20 1.78839270771153
60   B        Tim      20 5.07140678136995
61   C        Bob      21 3.47818616035335
62   C      Nancy      21 4.28526474048439
63   C        Tim      21 4.22597602946575
64   A        Bob      22 1.91700925257901
65   A      Nancy      22 2.96317997587458
66   A        Tim      22 2.53506974227672
67   B        Bob      23 5.52714403395316
68   B      Nancy      23  3.3618513551059
69   B        Tim      23 4.85869007113978
70   C        Bob      24  3.4367068543959
71   C      Nancy      24 4.47769879000349
72   C        Tim      24 5.77340483757836
73   A        Bob      25 4.78524317734622
74   A      Nancy      25 3.55373702554664
75   A        Tim      25 2.88541465503637
76   B        Bob      26 4.62885302019139
77   B      Nancy      26 3.59430293369092
78   B        Tim      26 2.29610255924296
79   C        Bob      27 4.38433001299722
80   C      Nancy      27 3.77825207859976
81   C        Tim      27 2.12163194694365

How do I pull out 2 of each trt x individual combinations with a unique session number? This is an example what I want the dataframe to look like:

       trt individual session             data
    1    A        Bob       1 3.72013685581385
    5    B      Nancy       2 3.43615965145765
    7    C        Bob       3 4.36315054477571
    12   A        Tim       4 3.09084920968413
    15   B        Tim       5 3.40778069131207
    17   C      Nancy       6 0.14578811080644
    19   A        Bob       7 3.69131009783284
    29   A      Nancy      10 5.17300616827746
    31   B        Bob      11 4.50892032922527
    34   C        Bob      12 4.54149796547394
    39   A        Tim      13 3.02932375225388
    40   B        Bob      14 3.59660910672907
    47   A      Nancy      16 2.19932140340504
    51   B        Tim      17 4.02576463486266
    54   C        Tim      18  3.2024114782253
    59   B      Nancy      20 1.78839270771153
    71   C      Nancy      24 4.47769879000349
    81   C        Tim      27 2.12163194694365

I have tried a couple things with no luck.

I have tried to just randomly select two trt x individual combinations, but I end up with duplicate session values:

setDT((df))
df[ , .SD[sample(.N, 2)] , keyby = .(trt, individual)]
    trt individual session             data
 1:   A        Bob      25  2.7560788894668
 2:   A        Bob      19 4.12040841647523
 3:   A      Nancy       4 5.35362338127901
 4:   A      Nancy      19 5.51636882737692
 5:   A        Tim      19 5.10553640201998
 6:   A        Tim       1 2.77380671625473
 7:   B        Bob      23 3.50585105164409
 8:   B        Bob       8 3.58167259470814
 9:   B      Nancy      23 2.85301307507985
10:   B      Nancy       8 2.85179395539781
11:   B        Tim      26 2.40666507132474
12:   B        Tim      20 3.31276311351286
13:   C        Bob      24 3.19076007024549
14:   C        Bob       3 3.59146613276121
15:   C      Nancy       9 4.46606667880457
16:   C      Nancy      15 2.25405252536256
17:   C        Tim      12 4.43111661206133
18:   C        Tim      27 4.23868848646589

I have tried randomly selecting one of each session number and then pulling 2 trt x individual combinations, but it typically comes back with an error since the random selection doesnt grab an equal number of trt x individual combinations:

ind <- sapply( unique(df$session ) , function(x) sample( which(df$session == x) , 1) )
df.unique <- df[ind, ]
df.sub <- df.unique[, .SD[sample(.N, 2)] , by = .(trt, individual)]
Error in `[.data.frame`(df.unique, , .SD[sample(.N, 2)], by = .(trt, individual)) : 
  unused argument (by = .(trt, individual))

Thanks in advance for your help!

解决方案

Perhaps there is a clever way to sample, but here's a straightforward idea to get you started in the meanwhile:

setDT(df)
setkey(df, session)

usedsessions = 0 # some value that's not a session number
df[, {
       res = .SD[!.(usedsessions)][sample(.N, 2)]
       usedsessions = c(usedsessions, res$session)
       res
     }
   , by = .(trt, individual)]
#    trt individual session     data
# 1:   A        Bob       7 4.256668
# 2:   A        Bob      25 2.431821
# 3:   A      Nancy      16 4.785859
# 4:   A      Nancy      19 4.865248
# 5:   A        Tim       4 3.303689
# 6:   A        Tim      13 3.550261
# 7:   B        Bob      26 3.987136
# 8:   B        Bob      17 3.283055
# 9:   B      Nancy      14 3.177226
#10:   B      Nancy       2 3.639542
#11:   B        Tim       8 2.168447
#12:   B        Tim       5 3.521123
#13:   C        Bob      21 3.284245
#14:   C        Bob      12 5.773098
#15:   C      Nancy      24 4.624428
#16:   C      Nancy       9 3.235467
#17:   C        Tim      18 4.001395
#18:   C        Tim      27 5.002110

You'll probably need to add corner case processing (e.g. if there is no such sampling).

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