为什么％＆gt;％不适用于apply（）？

Question

考虑以下数据：

library(dplyr)
df <- structure(list(Total = c(3450, 1728, 122, 5300), 
                   A1 = c(1092, 497, 4, 1593), 
                   A2 = c(596, 156, 29, 781), 
                   A3 = c(801, 417, 36, 1254), 
                   A4 = c(107, 11, 0, 118), 
                   A5 = c(614, 217, 21, 852), 
                   A6 = c(132, 47, 0, 179), 
                   A7 = c(108, 383, 32, 523)), 
              .Names = c("Total", paste0("A", 1:7)), 
              row.names = c(paste0("B", 1:3), "Total"), 
              class = "data.frame")

      Total   A1  A2   A3  A4  A5  A6  A7
B1     3450 1092 596  801 107 614 132 108
B2     1728  497 156  417  11 217  47 383
B3      122    4  29   36   0  21   0  32
Total  5300 1593 781 1254 118 852 179 523

问题如下：让行Total为分母，计算同一列中每个单元格与行Total的百分比。

在这种情况下，我会做

df_names <- rownames(df)

df2 <- apply(
  round(
    sweep(df, 
          STATS = as.matrix(df[rownames(df) == "Total",]),
          MARGIN = 2,
          FUN = "/") * 100,
    digits = 2),
  MARGIN = 2,
  FUN = paste0,
  "%")

rownames(df2) <- df_names
df2 <- as.data.frame(df2)

       Total     A1     A2     A3     A4     A5     A6     A7
B1    65.09% 68.55% 76.31% 63.88% 90.68% 72.07% 73.74% 20.65%
B2     32.6%  31.2% 19.97% 33.25%  9.32% 25.47% 26.26% 73.23%
B3      2.3%  0.25%  3.71%  2.87%     0%  2.46%     0%  6.12%
Total   100%   100%   100%   100%   100%   100%   100%   100%

但，虽然这有效，但我更喜欢使用%>%：

df3 <- sweep(df, 
             STATS = as.matrix(df[rownames(df) == "Total",]),
             MARGIN = 2,
             FUN = "/") * 100 %>%
       round(digits = 2) %>%
       apply(., MARGIN = 2, FUN = paste0, "%")

Error in apply(., MARGIN = 2, FUN = paste0, "%") : 
  dim(X) must have a positive length

为什么使用%>%的上述方法不起作用？

Answer 1

问题是尾随数字100以及它如何进入管道。使用()和apply的环绕可以正常工作，或者用于较少参与的解决方案：

(sweep(df, 
      STATS = as.matrix(df[rownames(df) == "Total",]),
      MARGIN = 2,
      FUN = "/") * 100) %>% round(digits = 2) %>%
    mutate_all(function(x)sprintf('%s%%', x))

   Total     A1     A2     A3     A4     A5     A6     A7
1 65.09% 68.55% 76.31% 63.88% 90.68% 72.07% 73.74% 20.65%
2  32.6%  31.2% 19.97% 33.25%  9.32% 25.47% 26.26% 73.23%
3   2.3%  0.25%  3.71%  2.87%     0%  2.46%     0%  6.12%
4   100%   100%   100%   100%   100%   100%   100%   100%

编辑以显示一些原因......

%>%需要两个参数，并生成一个环境。参数为lhs和rhs，左侧为输入值

sweep(df, 
      STATS = as.matrix(df[rownames(df) == "Total",]),
      MARGIN = 2,
      FUN = "/") * 100 %>% print

[1] 100
           Total          A1         A2         A3         A4         A5        A6         A7
B1     65.094340  68.5499058  76.312420  63.875598  90.677966  72.065728  73.74302  20.650096
B2     32.603774  31.1989956  19.974392  33.253589   9.322034  25.469484  26.25698  73.231358
B3      2.301887   0.2510986   3.713188   2.870813   0.000000   2.464789   0.00000   6.118547
Total 100.000000 100.0000000 100.000000 100.000000 100.000000 100.000000 100.00000 100.000000

注意[1] 100

现在有了parens

(sweep(df, 
      STATS = as.matrix(df[rownames(df) == "Total",]),
      MARGIN = 2,
      FUN = "/") * 100) %>% print

           Total          A1         A2         A3         A4         A5        A6         A7
B1     65.094340  68.5499058  76.312420  63.875598  90.677966  72.065728  73.74302  20.650096
B2     32.603774  31.1989956  19.974392  33.253589   9.322034  25.469484  26.25698  73.231358
B3      2.301887   0.2510986   3.713188   2.870813   0.000000   2.464789   0.00000   6.118547
Total 100.000000 100.0000000 100.000000 100.000000 100.000000 100.000000 100.00000 100.000000

括号也有效

sweep(df, 
      STATS = as.matrix(df[rownames(df) == "Total",]),
      MARGIN = 2,
      FUN = "/") %>% {
          round(. * 100, 2) %>% mutate_all(function(x)sprintf('%s%%',x))
    }

   Total     A1     A2     A3     A4     A5     A6     A7
1 65.09% 68.55% 76.31% 63.88% 90.68% 72.07% 73.74% 20.65%
2  32.6%  31.2% 19.97% 33.25%  9.32% 25.47% 26.26% 73.23%
3   2.3%  0.25%  3.71%  2.87%     0%  2.46%     0%  6.12%
4   100%   100%   100%   100%   100%   100%   100%   100%