使用 tidyverse 計算百分比

Question

我希望將 Type 和 Stream_Order 中每個組的計數除以所有組的總計數。 我有下面的數據框。 例如，對於“stoich”組中流序“1”的沉積物樣本，我想將“n”值 1445 除以“all”組中的對應值：(1445/6312)*100 = 22.89 到計算一個百分比。 到目前為止我有

perc %>%
  group_by(Type,Stream_Order) %>%
  mutate(percentage = n/total*100)

但我意識到我錯過了很多。 當然我可以手動完成這些，但我希望將來為更大的數據集做好准備。

> dput(perc)
structure(list(Type = c("Sediment", "Sediment", "Surface Water", 
"Surface Water", "Sediment", "Sediment", "Surface Water", "Surface Water", 
"Sediment", "Sediment", "Surface Water", "Surface Water", "Sediment", 
"Sediment", "Surface Water", "Surface Water"), Stream_Order = structure(c(1L, 
2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L), .Label = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "Not_Provided"), class = "factor"), 
    n = c(1445L, 639L, 1736L, 815L, 4580L, 2456L, 3999L, 1743L, 
    287L, 109L, 656L, 450L, 6312L, 3204L, 6391L, 3008L), meancorr = c(0.292071913949931, 
    0.262339039457943, -0.089326777121153, -0.259460896530476, 
    7.82761635773663, 8.66618572293474, 9.85787757997926, 8.06533374329178, 
    -3.04777412402461, -3.0807907346078, -3.07173828840263, -3.17877354374679, 
    5.60801895761982, 6.59050580264329, 5.82874680645636, 4.12764893257951
    ), max = c(1.99996435865152, 1.98765952360616, 1.99718852523326, 
    1.99609913106005, 32.9381587597398, 33.1286964665043, 38.3287173847001, 
    28.8178499883634, -2.00562491853763, -2.01247690847229, -2.00017475758744, 
    -2.00391643807592, 32.9381587597398, 33.1286964665043, 38.3287173847001, 
    28.8178499883634), min = c(-1.99674391114338, -1.99895657727752, 
    -1.99815485165027, -1.9986527287077, 2.00180602762732, 2.00187366008464, 
    2.01245579505433, 2.0066032552998, -7.194006200101, -6.68838345991404, 
    -6.69361085662866, -6.27066223419871, -7.194006200101, -6.68838345991404, 
    -6.69361085662866, -6.27066223419871), Group = c("stoch", 
    "stoch", "stoch", "stoch", "hetero", "hetero", "hetero", 
    "hetero", "homo", "homo", "homo", "homo", "all", "all", "all", 
    "all")), class = "data.frame", row.names = c(NA, -16L))

Answer 1

我們可以使用 'Group' 列上的邏輯表達式（ Group == 'all' ）子集 'n'

library(dplyr)
perc <- perc %>%
  group_by(Type, Stream_Order) %>%
  mutate(percentage = n/n[Group == 'all']*100) %>%
  ungroup

-輸出

perc
# A tibble: 16 × 8
   Type          Stream_Order     n meancorr   max   min Group  percentage
   <chr>         <fct>        <int>    <dbl> <dbl> <dbl> <chr>       <dbl>
 1 Sediment      1             1445   0.292   2.00 -2.00 stoch       22.9 
 2 Sediment      2              639   0.262   1.99 -2.00 stoch       19.9 
 3 Surface Water 1             1736  -0.0893  2.00 -2.00 stoch       27.2 
 4 Surface Water 2              815  -0.259   2.00 -2.00 stoch       27.1 
 5 Sediment      1             4580   7.83   32.9   2.00 hetero      72.6 
 6 Sediment      2             2456   8.67   33.1   2.00 hetero      76.7 
 7 Surface Water 1             3999   9.86   38.3   2.01 hetero      62.6 
 8 Surface Water 2             1743   8.07   28.8   2.01 hetero      57.9 
 9 Sediment      1              287  -3.05   -2.01 -7.19 homo         4.55
10 Sediment      2              109  -3.08   -2.01 -6.69 homo         3.40
11 Surface Water 1              656  -3.07   -2.00 -6.69 homo        10.3 
12 Surface Water 2              450  -3.18   -2.00 -6.27 homo        15.0 
13 Sediment      1             6312   5.61   32.9  -7.19 all        100   
14 Sediment      2             3204   6.59   33.1  -6.69 all        100   
15 Surface Water 1             6391   5.83   38.3  -6.69 all        100   
16 Surface Water 2             3008   4.13   28.8  -6.27 all        100