按更大的组订购DataFrame (R)

Question

我想按一个组的'sum‘对一个数据帧进行排序。因此，我不希望数据帧按组排序，而是按组的总量排序。也就是说，我想知道哪一组是最大的，1、2或3，然后根据它对值进行排序。所以，假设组3是最大的组，那么我希望组3在顶部，我希望组3的值按降序排列。

set.seed(123)
d <- data.frame(
  x   = runif(90),
  grp = gl(3, 30))

谢谢!

Answer 1

使用dplyr的一种方法是

d %>% 
  group_by(grp) %>% 
  mutate(sum_ = sum(x)) %>% 
  arrange(desc(sum_), desc(x)) %>% 
  select(-sum_)

基本上，我们创建一个临时变量sum_来表示每个组的x的总和，然后首先按照sum_排列，然后按照x排列。之后，我们删除sum_，因为不再需要它。

输出

# A tibble: 90 x 2
# Groups:   grp [3]
#        x grp  
#    <dbl> <fct>
#  1 0.994 1    
#  2 0.957 1    
#  3 0.955 1    
#  4 0.940 1    
#  5 0.900 1    
#  6 0.892 1    
#  7 0.890 1    
#  8 0.883 1    
#  9 0.788 1    
# 10 0.709 1    
# ... with 80 more rows

Answer 2

在base中，你可以先使用order on all，然后用grp拆分，再用order of rowsum拆分rbind。

d <- d[order(-d$x),]
do.call(rbind, split(d, d$grp)[order(-rowsum(d$x, d$grp))])
#                x grp
#1.24 0.9942697766   1
#1.11 0.9568333453   1
#1.20 0.9545036491   1
#1.5  0.9404672843   1
#1.16 0.8998249704   1
#1.8  0.8924190444   1
#1.21 0.8895393161   1
#1.4  0.8830174040   1
#1.2  0.7883051354   1
#1.26 0.7085304682   1
#1.22 0.6928034062   1
#1.13 0.6775706355   1
#1.25 0.6557057991   1
#1.23 0.6405068138   1
#1.28 0.5941420204   1
#1.14 0.5726334020   1
#1.9  0.5514350145   1
#1.27 0.5440660247   1
#1.7  0.5281054880   1
#1.10 0.4566147353   1
#1.12 0.4533341562   1
#1.3  0.4089769218   1
#1.19 0.3279207193   1
#1.29 0.2891597373   1
#1.1  0.2875775201   1
#1.17 0.2460877344   1
#1.30 0.1471136473   1
#1.15 0.1029246827   1
#1.6  0.0455564994   1
#1.18 0.0420595335   1
#3.87 0.9849569800   3
#3.88 0.8930511144   3
#3.89 0.8864690608   3
#3.65 0.8146400389   3
#3.68 0.8123895095   3
#3.67 0.8100643530   3
#3.69 0.7943423211   3
#3.84 0.7881958340   3
#3.71 0.7544751586   3
#3.73 0.7101824014   3
#3.82 0.6680555874   3
#3.61 0.6651151946   3
#3.72 0.6292211316   3
#3.78 0.6127710033   3
#3.75 0.4753165741   3
#3.66 0.4485163414   3
#3.70 0.4398316876   3
#3.86 0.4348927415   3
#3.83 0.4176467797   3
#3.63 0.3839696378   3
#3.77 0.3798165377   3
#3.79 0.3517979092   3
#3.64 0.2743836446   3
#3.81 0.2436194727   3
#3.76 0.2201188852   3
#3.90 0.1750526503   3
#3.80 0.1111354243   3
#3.85 0.1028646443   3
#3.62 0.0948406609   3
#3.74 0.0006247733   3
#2.31 0.9630242325   2
#2.32 0.9022990451   2
#2.59 0.8950453592   2
#2.50 0.8578277153   2
#2.53 0.7989248456   2
#2.34 0.7954674177   2
#2.37 0.7584595375   2
#2.58 0.7533078643   2
#2.33 0.6907052784   2
#2.55 0.5609479838   2
#2.36 0.4777959711   2
#2.48 0.4659624503   2
#2.52 0.4422000742   2
#2.42 0.4145463358   2
#2.43 0.4137243263   2
#2.60 0.3744627759   2
#2.44 0.3688454509   2
#2.39 0.3181810076   2
#2.49 0.2659726404   2
#2.47 0.2330340995   2
#2.40 0.2316257854   2
#2.38 0.2164079358   2
#2.56 0.2065313896   2
#2.45 0.1524447477   2
#2.41 0.1428000224   2
#2.46 0.1388060634   2
#2.57 0.1275316502   2
#2.54 0.1218992600   2
#2.51 0.0458311667   2
#2.35 0.0246136845   2

另一种选择是重新排序grp的levels，然后可以在order中使用它。

d$grp <- factor(d$grp, levels(d$grp)[order(-rowsum(d$x, d$grp))])
d[order(d$grp, -d$x),]