用dplyr汇总后如何执行计算？

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篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了用dplyr汇总后如何执行计算？相关的知识，希望对你有一定的参考价值。

我已经通过以下方式对两组进行了逐组计算：

group_stats <- fev %>%
  group_by(smoking) %>%
  summarize(mean = mean(fev), n = n(), sd = sd(fev), se = sd / sqrt(n))

将产生具有两行的数据帧。现在我要计算以下内容：

  1. 两行之间的均值之差
  2. 类似于se的新se[row_1] ^ 2 + se[row_2] ^ 2值。

是否有整齐的方式？

[dput输出：

structure(list(age = c(8, 13, 8, 11, 8, 8, 9, 11, 15, 8, 6, 8, 
12, 8, 7, 9, 10, 8, 12, 13, 11, 12, 9, 7, 11, 14, 12, 9, 10, 
8, 10, 11, 6, 9, 11, 14, 11, 10, 9, 11, 11, 11, 8, 15, 7, 9, 
11, 11, 8, 8, 9, 11, 8, 11, 8, 8, 9, 14, 9, 10, 16, 13, 16, 10, 
8, 12, 9, 11, 8, 8, 12, 7, 13, 9, 7, 8, 10, 7, 13, 13, 8, 8, 
8, 16, 12, 10, 14, 9, 8, 7, 9, 11, 12, 13, 13, 10, 6, 13, 9, 
7, 6, 8, 6, 12, 11, 10, 11, 9, 10, 9, 13, 10, 12, 6, 13, 13, 
10, 11, 11, 8, 9, 9, 15, 7, 8, 15, 15, 8, 14, 10, 6, 10, 8, 6, 
12, 8, 13, 9, 16, 10, 11, 8, 6, 11, 10, 13, 12, 11, 10, 8, 8, 
10, 12, 10, 7, 9, 9, 12, 8, 7, 8, 9, 6, 11, 13, 8, 10, 12, 9, 
11, 13, 6, 13, 10, 8, 12, 6, 9, 9, 13, 11, 15, 6, 11, 10, 10, 
10, 7, 7, 12, 12, 10, 13, 10, 10, 10, 11, 10, 13, 9, 11, 12, 
10, 8, 11, 11, 10, 13, 13, 9, 8, 11, 9, 7, 10, 11, 9, 9, 11, 
14, 9, 7, 10, 7, 15, 14, 11, 9, 8, 7, 13, 13, 9, 9, 7, 11, 11, 
16, 9, 15, 7, 12, 7, 6, 8, 12, 8, 14, 9, 9, 8, 15, 10, 12, 13, 
11, 8, 11, 8, 11, 13, 8, 11, 9, 6, 10, 11, 10, 11, 9, 8, 12, 
12, 12, 8, 9, 15, 12, 14, 14, 9, 9, 10, 13, 8, 13, 13, 9, 9, 
8, 11, 9, 7, 7, 9, 9, 9, 9, 7, 8, 9, 15, 16, 10, 6, 9, 12, 9, 
8, 8, 8, 9, 11, 7, 15, 7, 11, 12, 8, 11, 14, 7, 6, 8, 7, 12, 
11, 13, 15, 14, 12, 8, 11, 10, 11, 9, 7, 9, 8, 10, 7, 8, 15, 
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6, 9, 10, 12, 13, 8, 7, 12, 11, 8, 11, 6, 11, 12, 7, 13, 7, 14, 
7, 10, 12, 10, 6, 9, 9, 9, 12, 8, 9, 6, 12, 7, 10, 9, 10, 11, 
11, 7, 10, 7, 10, 10, 16, 6, 6, 6, 10, 14, 12, 7, 10, 13, 11, 
9, 7, 7, 7, 11, 9, 7, 8, 11, 8, 8, 10, 10, 9, 8, 9, 12, 6, 8, 
7, 9, 11, 11, 9, 10, 9, 14, 9, 7, 8, 11, 7, 6, 12, 9, 8, 10, 
13, 10, 13, 15, 8, 9, 10, 15, 11, 9, 9, 13, 14, 6, 6, 10, 10, 
12, 14, 7, 11, 11, 13, 10, 7, 14, 10, 9, 9, 6, 14, 12, 10, 11, 
8, 12, 9, 11, 8, 16, 7, 14, 12, 13, 12, 8, 10, 11, 16, 9, 12, 
13, 10, 10, 11, 10, 10, 13, 9, 9, 14, 10, 9, 16, 9, 6, 11, 6, 
11, 12, 14, 15, 11, 11, 12, 12, 9, 8, 8, 11, 6, 9, 7, 10, 12, 
10, 7, 10, 8, 12, 9, 13, 12, 9, 8, 12, 11, 11, 8, 11, 11, 14, 
9, 16, 9, 6, 10, 10, 12, 15, 11, 8, 15, 8, 9, 7, 9, 10, 9, 8, 
10, 12, 7, 8, 11, 12, 10, 10, 8, 11, 8), fev = c(1.94, 3.785, 
2.123, 2.993, 2.094, 2.187, 2.23, 2.14, 5.793, 2.069, 1.979, 
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1.75, 2.09, 2.341, 2.371, 4.448, 3.208, 1.512, 2.175, 2.5, 2.795, 
2.752, 2.352, 4.683, 2.964, 2.503, 2.55, 2.923, 2.735, 2.241, 
4.336, 3.255, 2.642, 2.262, 3.06, 3, 1.905, 1.691, 2.476, 1.796, 
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4.007, 1.987, 4.872, 2.111, 3.074, 3.751, 3.193, 2.913, 1.999, 
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2.465, 2.293, 3.731, 2.288, 2.056, 1.877, 2.688, 2.52, 3.556, 
2.004, 3.166, 2.822, 1.932, 1.724, 3.111, 2.866, 1.811, 2.608, 
2.211, 2.754, 2.435), gender = structure(c(2L, 1L, 2L, 2L, 2L, 
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1L, 1L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 
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1L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 
2L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 
2L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 
1L, 1L, 1L, 1L, 2L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 
2L), .Label = c("f", "m"), class = "factor"), smoking = structure(c(1L, 
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1L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 
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1L, 1L, 1L, 1L, 1L), .Label = c("ns", "s"), class = "factor")), class = c("spec_tbl_df", 
"tbl_df", "tbl", "data.frame"), row.names = c(NA, -598L))
答案

在dplyr中，可以通过在[]中插入单个或范围值来引用每一行中的值。例如，mean列的第一个值可以通过mean[1]获得，第二个可以通过mean[2]获得，都可以通过mean[1:2]或mean[c(1, 2)]等获得。]

如果只有两行，您还可以通过last(mean)访问第二行，通过first(mean)访问第一行。

当您以前有group_by语句时，这些索引引用每个组的行索引。

但是，执行此操作的最佳方法是仅执行基本操作。例如，您可以计算指数，然后将其sum代入se，而对于diff，只需采用mean函数即可。

见下文：

group_stats <- fev %>%
  group_by(smoking) %>%
  summarize(mean = mean(fev), n = n(), sd = sd(fev), se = sd / sqrt(n)) %>%
  mutate(se_sum = sum(se ^ 2),
         se_idx = se[1] ^ 2 + se[2] ^ 2,
         mean_diff = diff(mean),
         mean_idx = mean[2] - mean[1],
         mean_diffLast = last(mean) - first(mean))

输出：
group_stats

# A tibble: 2 x 10
  smoking  mean     n    sd     se se_sum se_idx mean_diff mean_idx mean_diffLast
  <fct>   <dbl> <int> <dbl>  <dbl>  <dbl>  <dbl>     <dbl>    <dbl>         <dbl>
1 ns       2.61   539 0.788 0.0339 0.0111 0.0111     0.628    0.628         0.628
2 s        3.24    59 0.764 0.0995 0.0111 0.0111     0.628    0.628         0.628

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