如何找到R数据帧的列值之和？
R数据框包含可能代表相似类型变量的列；因此，我们可能希望找到每个列的值的总和，然后基于该总和进行比较。这可以借助sum函数来完成，但是首先我们需要提取列以找到和。
示例
请看以下数据帧-
> set.seed(1) > x1<-rnorm(20) > x2<-rnorm(20,0.5) > x3<-rnorm(20,1) > x4<-rnorm(20,1.5) > x5<-rnorm(20,2) > x6<-rnorm(20,2.5) > x7<-rnorm(20,3) > df<-data.frame(x1,x2,x3,x4,x5,x6,x7) > df x1 x2 x3 x4 x5 x6 x7 1 -0.62645381 1.41897737 0.83547640 3.9016178 1.4313313 1.879633 2.494043 2 0.18364332 1.28213630 0.74663832 1.4607600 1.8648214 2.542116 4.343039 3 -0.83562861 0.57456498 1.69696338 2.1897394 3.1780870 1.589078 2.785421 4 1.59528080 -1.48935170 1.55666320 1.5280022 0.4764332 2.658029 2.820443 5 0.32950777 1.11982575 0.31124431 0.7567268 2.5939462 1.845415 2.899809 6 -0.82046838 0.44387126 0.29250484 1.6887923 2.3329504 4.267287 3.712666 7 0.48742905 0.34420449 1.36458196 -0.3049586 3.0630998 3.216707 2.926436 8 0.73832471 -0.97075238 1.76853292 2.9655549 1.6958161 3.410174 2.962366 9 0.57578135 0.02184994 0.88765379 1.6532533 2.3700188 2.884185 2.318340 10 -0.30538839 0.91794156 1.88110773 3.6726117 2.2670988 4.182176 2.675730 11 1.51178117 1.85867955 1.39810588 1.9755095 1.4574800 1.864264 3.060160 12 0.38984324 0.39721227 0.38797361 0.7900536 3.2078678 2.038355 2.411106 13 -0.62124058 0.88767161 1.34111969 2.1107264 3.1604026 3.932282 3.531496 14 -2.21469989 0.44619496 -0.12936310 0.5659024 2.7002136 1.849304 1.481606 15 1.12493092 -0.87705956 2.43302370 0.2463666 3.5868335 2.292619 3.306558 16 -0.04493361 0.08500544 2.98039990 1.7914462 2.5584864 2.107192 1.463550 17 -0.01619026 0.10571005 0.63277852 1.0567081 0.7234078 2.180007 2.699024 18 0.94383621 0.44068660 -0.04413463 1.5011054 1.4267346 2.220887 2.471720 19 0.82122120 1.60002537 1.56971963 1.5743413 0.7753874 2.994188 2.347905 20 0.59390132 1.26317575 0.86494540 0.9104791 1.5265994 2.322670 2.943103
找出所有列的总和-
> sum(df$x1) [1] 3.810478 > sum(df$x2) [1] 9.87057 > sum(df$x3) [1] 22.77594 > sum(df$x4) [1] 32.03474 > sum(df$x5) [1] 42.39702 > sum(df$x6) [1] 52.27657 > sum(df$x7) [1] 55.65452
让我们再看一个例子-
> y1<-1:10 > y2<-11:20 > y3<-rep(c(1,2,3,4,5),times=2) > df_y<-data.frame(y1,y2,y3) > df_y y1 y2 y3 1 1 11 1 2 2 12 2 3 3 13 3 4 4 14 4 5 5 15 5 6 6 16 1 7 7 17 2 8 8 18 3 9 9 19 4 10 10 20 5 > sum(df_y$y1) [1] 55 > sum(df_y$y2) [1] 155 > sum(df_y$y3) [1] 30