如何使用 group_by 汇总所有列并进行汇总？答案

【问题标题】：How to summarise all columns using group_by and summarise?如何使用 group_by 汇总所有列并进行汇总？
【发布时间】：2019-07-30 00:04:15
【问题描述】：

我正在尝试整理我的日常活动数据（加速度计数据）。我想总结所有列每天重复的行。我有 32 行（有些是重复的）和 90 列（一个主题的数据）。

# Example of my data with 32 rows and 14 columns

df <- data.frame(LbNr = c(22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002,22002),
Type = c("A2. Working" ,"A1. NonWorking" ,"A1. NonWorking" ,"A4. SleepWeek" ,"A1. NonWorking" ,"A2. Working" ,"A1. NonWorking" ,"A4. SleepWeek" ,"A4. SleepWeek" ,"A1. NonWorking" ,"A2. Working" ,"A1. NonWorking" ,"A1. NonWorking" ,"A4. SleepWeek" ,"A1. NonWorking" ,"A2. Working" ,"A1. NonWorking" ,"A4. SleepWeek" ,"A4. SleepWeek" ,"A1. NonWorking" ,"A2. Working" ,"A1. NonWorking" ,"A1. NonWorking" ,"C4. SleepWeekend" ,"C0. Leisure" ,"C0. Leisure" ,"C4. SleepWeekend" ,"C0. Leisure" ,"C4. SleepWeekend" ,"C4. SleepWeekend" ,"A1. NonWorking" ,"A2. Working"),
Weekday = c(1,1,2,2,2,2,2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,6,6,6,7,7,7,7,1,1,1),
Time = c(0.66667,5.66667,0.35,6.15,1.5,9.83333,6.05,0.11667,6.83333,1.33333,9.83333,6,0.03333,7.2,6.43333,5,5.23333,0.1,6.41667,0.96667,11.01667,5.6,0.43333,7.9,15.66667,0.03333,7.91667,15.61667,0.43333,6.33333,0.66667,6.83333),
lie = c(0.00583,0.37778,0.03556,4.84389,0.05444,0.05972,0.67639,0.0125,5.68806,0.02333,0.65278,0.23889,0.00917,7.2,0.45472,0.38333,0.29694,0.08,5.48694,0.01889,0.01028,0.12139,0.01694,6.96028,0.24472,0.00333,6.93639,0.11833,0.41861,5.74889,0.00861,0.07333),
sit = c(0.31194,4.36167,0.14417,1.30611,0.45083,6.64111,4.14306,0.10417,1.14528,0.51167,5.79417,3.11833,0,0,2.23944,2.79722,3.66583,0.00472,0.92972,0.29917,6.76806,4.21056,0.30222,0.92194,9.77694,0.00417,0.91833,12.02972,0.01472,0.58444,0.15806,5.58694),
stand = c(0.13389,0.47111,0.09139,0,0.67278,1.63667,0.51806,0,0,0.46417,1.81917,1.57472,0.01889,0,1.88917,0.88639,0.63028,0.00667,0,0.3975,1.83417,0.72528,0.05889,0.00667,2.33944,0.01361,0.03639,1.78139,0,0,0.25472,0.41167),
move = c(0.09056,0.34444,0.05167,0,0.21611,0.59472,0.34306,0,0,0.21333,0.525,0.72806,0.00528,0,0.76583,0.39194,0.41861,0.00667,0,0.14056,1.04694,0.36944,0.03778,0.00806,2.44583,0.00944,0.02083,0.93083,0,0,0.15417,0.235),
walk = c(0.11528,0.10722,0.02722,0,0.10583,0.84194,0.35639,0,0,0.11694,1.00806,0.33167,0,0,1.04611,0.51389,0.20833,0,0,0.09333,1.28528,0.16083,0.0175,0.00306,0.79972,0.00278,0.00472,0.65306,0,0,0.08139,0.49528),
run = c(0,0.00111,0,0,0,0.00167,0.00194,0,0,0,0.00083,0.00083,0,0,0.00333,0.0025,0.00083,0,0,0.00139,0.00472,0,0,0,0.00194,0,0,0.08694,0,0,0,0.00111),
stairs = c(0.00917,0.00333,0,0,0,0.0575,0.01111,0,0,0.00389,0.03333,0.0075,0,0,0.03472,0.02472,0.00472,0.00194,0,0.00583,0.06722,0.0125,0,0,0.05806,0,0,0.01639,0,0,0.00417,0.03),
cycle = c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.00778,0,0,0.01,0,0,0,0,0,0,0,0,0,0,0.00556,0),
WalkSlow = c(0.01222,0.02056,0.00389,0,0.03056,0.17417,0.03361,0,0,0.01889,0.35889,0.07778,0,0,0.07528,0.04222,0.03417,0,0,0.02444,0.13722,0.03361,0.00417,0,0.14,0,0.00056,0.08056,0,0,0.02278,0.08278),
WalkFast = c(0.10278,0.08639,0.02278,0,0.07417,0.66,0.32194,0,0,0.0975,0.64583,0.25139,0,0,0.97083,0.46861,0.17222,0,0,0.06861,1.14694,0.12667,0.01306,0.00278,0.65444,0.00194,0.0025,0.56944,0,0,0.0575,0.41))

我尝试了一些小代码，但几乎都失败了。下面的代码是我能得到的，它太大了。我想知道是否有任何其他方法可以做得更小。

# LbNr = subjects' id
# Weekday = 1 Monday.... 7 Sunday
# Type = activities: A1. NonWorking, A2. Working, A4. SleepWeek, C0. Leisure, C4. SleepWeekend

# code
df %>% select(LbNr, Type, Weekday, Time, lie:IncTrunkWalk) %>% 
  group_by(LbNr, Type, Weekday) %>% 
  summarise(n = n(), Time = sum(Time),lie   = sum(lie), sit = sum(sit), stand = sum(stand),
            move = sum(move),   walk = sum(walk), run = sum(run),   stairs = sum(stairs),
            cycle = sum(cycle), row = sum(row), WalkSlow = sum(WalkSlow),
            WalkFast = sum(WalkFast)) %>% 
  arrange(Weekday) %>% filter(Weekday %in% c('3':'7'))

到目前为止，这段代码还有一个问题。我的问题是星期六“6”，当我连接时间可能是星期六收到从星期五开始的活动（见下面的例子），有时会出现“A1.NonWorking”或“A4.SleepWeek”，取决于志愿者.我想总结一下“C0.休闲”这个不同的活动。如果可能的话，我想在一个代码中完成。

#   LbNr      Type           Weekday   n   Time    lie    sit
#   <dbl>    <fct>             <dbl> <int> <dbl>  <dbl>   <dbl>   
#8  22002 A2. Working            5     1   11.0   0.0103  6.77 
#9  22002 A4. SleepWeek          5     1   6.42   5.49    0.930  
#10 22002 A1. NonWorking         6     1   0.433  0.0169  0.302
#11 22002 C0. Leisure            6     1   15.7   0.245   9.78
#12 22002 C4. SleepWeekend       6     1   7.9    6.96    0.922
#13 22002 C0. Leisure            7     2   15.6   0.122   12.0



#I would like to get something like this.
#   LbNr      Type           Weekday   n   Time    lie    sit
#   <dbl>    <fct>             <dbl> <int> <dbl>  <dbl>   <dbl>   
#8  22002 A2. Working            5     1   11.0   0.0103  6.77 
#9  22002 A4. SleepWeek          5     1   6.42   5.49    0.930  
#10 22002 C0. Leisure            6     1   16.133 0.2619  10.082
#11 22002 C4. SleepWeekend       6     1   7.9    6.96    0.922
#12 22002 C0. Leisure            7     2   15.6   0.122   12.0

对于第一个问题，我希望得到一个小代码。此外，如果可能的话，我希望能够为周六不同活动的总和获得更好的代码。

提前致谢，路易斯

【问题讨论】：

在第一组代码中，您正在使用sum 执行一系列列，这是行不通的。二、为什么sum改成mean
我需要sum，我在这里看到的第二个代码，所以我试着看看我得到了什么。此外，我尝试将mean 更改为sum，但它不起作用。
我看到的一个潜在问题是在 group_by 和 mutate_at group_by(LbNr, Type, Weekday) %>% summarise_at(vars(LbNr:IncTrunkWalk), 中使用相同的变量。 LbNr
我可以看到。我在没有它的情况下检查了它，但效果也不好。 df %>% group_by(LbNr, Type, Weekday) %>% summarise_at(vars(Time:IncTrunkWalk))

标签： r dplyr

【解决方案1】：

如果没有更好的示例，很难尝试回答您的问题（即，您可以dput() 您的数据为我们提供示例）。但这是您上一个问题的解决方案：“对于第一个问题，我希望得到一个包含所有列的重复行总和的表。此外，如果可能的话，我希望得到一个更好的总和代码周六的不同活动。”

# create toy data of 3 different IDs, 3 different types, and repeated days
df <- data.frame(id=sample(c(1:3),100,T),
                 type=sample(letters[1:3],100,T),
                 day=sample(c(1:7),100,T),
                 matrix(runif(300),nrow=100),
                 stringsAsFactors = F)

# gather data, summarize each activity column by ID, type and day
# and select Saturday==6
df %>% gather(k,v,-id,-type,-day) %>% 
  group_by(id,type,day,k) %>% 
  summarise(sum=sum(v)) %>% 
  filter(day==6) %>% 
  spread(k,sum)

# A tibble: 8 x 6
# Groups:   id, type, day [8]
     id type    day    X1    X2    X3
  <int> <chr> <int> <dbl> <dbl> <dbl>
1     1 a         6 1.85  3.26  2.09 
2     1 b         6 0.604 0.583 0.586
3     1 c         6 0.163 0.663 0.624
4     2 a         6 0.185 0.952 0.349
5     2 b         6 1.16  0.832 0.974
6     2 c         6 0.906 1.62  0.853
7     3 b         6 0.671 1.39  0.887
8     3 c         6 0.449 0.150 0.647

更新
这是使用提供的新数据的更新解决方案。

df %>% group_by(LbNr,Type,Weekday) %>% summarise_all(.,sum)

# A tibble: 20 x 14
# Groups:   LbNr, Type [5]
    LbNr Type  Weekday   Time    lie     sit   stand    move    walk     run  stairs   cycle
   <dbl> <fct>   <dbl>  <dbl>  <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>
 1 22002 A1. ~       1  6.33  0.386  4.52e+0 0.726   0.499   0.189   0.00111 0.0075  0.00556
 2 22002 A1. ~       2  7.9   0.766  4.74e+0 1.28    0.611   0.489   0.00194 0.0111  0      
 3 22002 A1. ~       3  7.33  0.262  3.63e+0 2.04    0.941   0.449   0.00083 0.0114  0      
 4 22002 A1. ~       4 11.7   0.761  5.91e+0 2.54    1.19    1.25    0.00416 0.0394  0.00778
 5 22002 A1. ~       5  6.57  0.140  4.51e+0 1.12    0.51    0.254   0.00139 0.0183  0.01   
 6 22002 A1. ~       6  0.433 0.0169 3.02e-1 0.0589  0.0378  0.0175  0       0       0      
 7 22002 A2. ~       1  7.5   0.0792 5.90e+0 0.546   0.326   0.611   0.00111 0.0392  0      
 8 22002 A2. ~       2  9.83  0.0597 6.64e+0 1.64    0.595   0.842   0.00167 0.0575  0      
 9 22002 A2. ~       3  9.83  0.653  5.79e+0 1.82    0.525   1.01    0.00083 0.0333  0      
10 22002 A2. ~       4  5     0.383  2.80e+0 0.886   0.392   0.514   0.0025  0.0247  0      
11 22002 A2. ~       5 11.0   0.0103 6.77e+0 1.83    1.05    1.29    0.00472 0.0672  0      
12 22002 A4. ~       2  6.27  4.86   1.41e+0 0       0       0       0       0       0      
13 22002 A4. ~       3  6.83  5.69   1.15e+0 0       0       0       0       0       0      
14 22002 A4. ~       4  7.3   7.28   4.72e-3 0.00667 0.00667 0       0       0.00194 0      
15 22002 A4. ~       5  6.42  5.49   9.30e-1 0       0       0       0       0       0      
16 22002 C0. ~       6 15.7   0.245  9.78e+0 2.34    2.45    0.800   0.00194 0.0581  0      
17 22002 C0. ~       7 15.6   0.122  1.20e+1 1.80    0.940   0.656   0.0869  0.0164  0      
18 22002 C4. ~       1  6.33  5.75   5.84e-1 0       0       0       0       0       0      
19 22002 C4. ~       6  7.9   6.96   9.22e-1 0.00667 0.00806 0.00306 0       0       0      
20 22002 C4. ~       7  8.35  7.36   9.33e-1 0.0364  0.0208  0.00472 0       0       0      
# ... with 2 more variables: WalkSlow <dbl>, WalkFast <dbl>

我认为这回答了您关于想要一个“小代码”的第一个问题。我仍然不明白你的第二个问题，即“我希望在周六获得更好的不同活动总和的代码。”这是否意味着您只想总结周六的不同活动（躺、坐等）？或者您想总结不同类型（A2、C0 等）的活动？

df %>% group_by(LbNr,Type,Weekday) %>% summarise_all(.,sum) %>% 
  filter(Weekday==6)

# A tibble: 3 x 14
# Groups:   LbNr, Type [3]
   LbNr Type  Weekday   Time    lie   sit   stand    move    walk     run stairs cycle WalkSlow
  <dbl> <fct>   <dbl>  <dbl>  <dbl> <dbl>   <dbl>   <dbl>   <dbl>   <dbl>  <dbl> <dbl>    <dbl>
1 22002 A1. ~       6  0.433 0.0169 0.302 0.0589  0.0378  0.0175  0       0          0  0.00417
2 22002 C0. ~       6 15.7   0.245  9.78  2.34    2.45    0.800   0.00194 0.0581     0  0.14   
3 22002 C4. ~       6  7.9   6.96   0.922 0.00667 0.00806 0.00306 0       0          0  0      
# ... with 1 more variable: WalkFast <dbl>

# summarise across different activities, for each column, on Saturday only
df %>% group_by(LbNr,Type,Weekday) %>% summarise_all(.,sum) %>% 
  filter(Weekday==6) %>% group_by(LbNr) %>% select(-Type,-Weekday) %>% 
  summarise_all(.,sum)

# A tibble: 1 x 12
   LbNr  Time   lie   sit stand  move  walk     run stairs cycle WalkSlow WalkFast
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>   <dbl>  <dbl> <dbl>    <dbl>    <dbl>
1 22002    24  7.22  11.0  2.41  2.49 0.820 0.00194 0.0581     0    0.144    0.670

【讨论】：

嗨@kstew，我做了你所做的，但我无法像你的例子那样得到它。关于第一个代码，我又写了一遍，但我认为这不是最聪明的写法。这是工作！！！但是代码太大了。此外，我需要使用您所做的代码来实现此代码。我用数据和新代码编辑了问题。期待您的回复，提前致谢。
嗨 Luiz，感谢您提供数据。但是当我尝试将其作为数据框读取时它不起作用，因此我无法使用您的数据尝试您的代码。您能否更具体地说明您的预期输出是什么？你能用我的代码展示你尝试过的东西吗？谢谢。
嗨@kstew。感谢您尝试帮助我，并为长时间的延迟感到抱歉。我更改了数据，也更改了代码以适应它，现在我认为您可以重现我的代码。此外，我给出了另一个我期望的输出示例。谢谢你的时间
嗨 Luiz，再次感谢您提供数据。我已经更新了我的答案。
嗨@kstew。谢谢你帮助我。你的第一个代码就是我需要的。