我有一个数据集，其中包含各种药物对不同细菌的敏感性。 我想获得生物体的敏感性频率。有没有办法简化这一过程，而不是对每种药物进行复制/粘贴？ 我正在考虑使用
applypacman::p_load(tidyverse,
janitor)
demo_dat <- data.frame(
stringsAsFactors = FALSE,
organism_name = c("Klebsiella pneumonia","Klebsiella pneumonia",
"Escherichia coli","Klebsiella pneumonia",
"Enterobacter cloacae","Escherichia coli",
"Klebsiella pneumonia","Escherichia coli",
"Escherichia coli","Escherichia coli",
"Klebsiella pneumonia","Klebsiella pneumonia",
"Escherichia coli","Klebsiella pneumonia",
"Escherichia coli","Serratia marcenscens",
"Klebsiella oxytoca","Escherichia coli",
"Proteus mirabilis","Escherichia coli"),
amox_clav_po = c("S",
"S","S","I","R","I","S","I","R","I",
"S","S","S","S","I","R","S","S","S",
"R"),
amp_sul_iv = c("S",
"I","S","S","R","R","S","S","R","I",
"S","I","S","I","R","R","S","S","S",
"R"),
cefaclor_po = c("S",
"S","S","S","R","S","S","S","S","S",
"S","S","S","S","R","R","S","S","S",
"S"),
ceftriaxone_iv = c("S",
"S","S","S","S","S","S","S","S","S",
"S","S","S","S","R","S","S","S","S",
"S")
)
demo_dat |>
group_by(organism_name) |>
summarise(susceptibility = sum((amox_clav_po == "S")/n()))
#> # A tibble: 6 × 2
#> organism_name susceptibility
#> <chr> <dbl>
#> 1 Enterobacter cloacae 0
#> 2 Escherichia coli 0.333
#> 3 Klebsiella oxytoca 1
#> 4 Klebsiella pneumonia 0.857
#> 5 Proteus mirabilis 1
#> 6 Serratia marcenscens 0
demo_dat |>
group_by(organism_name) |>
summarise(susceptibility = sum((amp_sul_iv == "S")/n()))
#> # A tibble: 6 × 2
#> organism_name susceptibility
#> <chr> <dbl>
#> 1 Enterobacter cloacae 0
#> 2 Escherichia coli 0.444
#> 3 Klebsiella oxytoca 1
#> 4 Klebsiella pneumonia 0.571
#> 5 Proteus mirabilis 1
#> 6 Serratia marcenscens 0
创建于 2024-01-29，使用 reprex v2.0.2
会议信息sessioninfo::session_info()
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您可以进行旋转和总结：
library(dplyr)
library(tidyr)
out <- demo_dat |>
pivot_longer(-organism_name) |>
summarize(
susceptibility = sum(value == "S")/n(),
.by = c(organism_name, name)
)
out
# # A tibble: 24 × 3
# organism_name name susceptibility
# <chr> <chr> <dbl>
# 1 Klebsiella pneumonia amox_clav_po 0.857
# 2 Klebsiella pneumonia amp_sul_iv 0.571
# 3 Klebsiella pneumonia cefaclor_po 1
# 4 Klebsiella pneumonia ceftriaxone_iv 1
# 5 Escherichia coli amox_clav_po 0.333
# 6 Escherichia coli amp_sul_iv 0.444
# 7 Escherichia coli cefaclor_po 0.889
# 8 Escherichia coli ceftriaxone_iv 0.889
# 9 Enterobacter cloacae amox_clav_po 0
# 10 Enterobacter cloacae amp_sul_iv 0
# # ℹ 14 more rows
# # ℹ Use `print(n = ...)` to see more rows
只是为了确保我们看到的内容与您的输出相同，
filter(out, name == "amox_clav_po")
# # A tibble: 6 × 3
# organism_name name susceptibility
# <chr> <chr> <dbl>
# 1 Klebsiella pneumonia amox_clav_po 0.857
# 2 Escherichia coli amox_clav_po 0.333
# 3 Enterobacter cloacae amox_clav_po 0
# 4 Serratia marcenscens amox_clav_po 0
# 5 Klebsiella oxytoca amox_clav_po 1
# 6 Proteus mirabilis amox_clav_po 1
您可以对数据中的所有值和列执行此操作。然后，如果您愿意，您可以选择那些具有
S使用基础R：
model.matrix(organism_name~0+., demo_dat)|>
cbind(demo_dat['organism_name']) |>
aggregate(.~organism_name, mean)
organism_name amox_clav_poI amox_clav_poR amox_clav_poS amp_sul_ivR amp_sul_ivS cefaclor_poS ceftriaxone_ivS
1 Enterobacter cloacae 0.0000000 1.0000000 0.0000000 1.0000000 0.0000000 0.0000000 1.0000000
2 Escherichia coli 0.4444444 0.2222222 0.3333333 0.4444444 0.4444444 0.8888889 0.8888889
3 Klebsiella oxytoca 0.0000000 0.0000000 1.0000000 0.0000000 1.0000000 1.0000000 1.0000000
4 Klebsiella pneumonia 0.1428571 0.0000000 0.8571429 0.0000000 0.5714286 1.0000000 1.0000000
5 Proteus mirabilis 0.0000000 0.0000000 1.0000000 0.0000000 1.0000000 1.0000000 1.0000000
6 Serratia marcenscens 0.0000000 1.0000000 0.0000000 1.0000000 0.0000000 0.0000000 1.0000000