英文:
Expanding values from multiple columns into binary values
问题
我有包含数字或NA值的列，我想重新排列它们，以使新列以不同的数字值命名，并且值为二进制的0或1值。
实际数据集包含许多其他变量，因此代码需要将函数细化到仅处理"stress"变量。
示例数据集：
df <- data.frame(stress1 = c('A', 'A', 'B', 'A'),stress2 = c(NA, 'B', 'C', 'B'),stress3 = c(NA, NA, NA, 'C'),stress4 = c(NA, NA, NA, 'D'))
期望的结果：
desiredoutcome <- data.frame (A = c(1, 1, 0, 1),B = c(0, 1, 1, 1),C = c(0, 0, 1, 1),D = c(0, 0, 0, 1))
英文:
I have columns with either numeric or NA values that I want to reorder so that the new columns are named with the distinct numeric values, and the values are binary 0 or 1 values.
The actual dataset contains many other variables, so the code would need to refine the function only to the 'stress' variables.
Example dataset:
df <- data.frame(stress1 = c('A', 'A', 'B', 'A'),stress2 = c(NA, 'B', 'C', 'B'),stress3 = c(NA, NA, NA, 'C'),stress4 = c(NA, NA, NA, 'D'))
Desired outcome:
desiredoutcome <- data.frame (A = c(1, 1, 0, 1),B = c(0, 1, 1, 1),C = c(0, 0, 1, 1),D = c(0, 0, 0, 1))
答案1
得分: 2
在 tidyr + dplyr 中：
library(tidyr)library(dplyr)df %>%mutate(id = row_number()) %>%pivot_longer(-id, values_drop_na = TRUE) %>%pivot_wider(names_from = "value", values_from = "name",values_fill = 0, values_fn = length)# id A B C D# 1 1 1 0 0 0# 2 2 1 1 0 0# 3 3 0 1 1 0# 4 4 1 1 1 1
或者在基本 R 中：
df$ID <- seq_along(df)table(cbind(df['ID'], unlist(df[1:4]))) |as.data.frame.matrix()# A B C D# 1 1 0 0 0# 2 1 1 0 0# 3 0 1 1 0# 4 1 1 1 1
英文:
In tidyr + dplyr.
library(tidyr)library(dplyr)df %>%mutate(id = row_number()) %>%pivot_longer(-id, values_drop_na = TRUE) %>%pivot_wider(names_from = "value", values_from = "name",values_fill = 0, values_fn = length)# id A B C D# 1 1 1 0 0 0# 2 2 1 1 0 0# 3 3 0 1 1 0# 4 4 1 1 1 1
Or in base R:
df$ID <- seq_along(df)table(cbind(df['ID'], unlist(df[1:4]))) |>as.data.frame.matrix()# A B C D# 1 1 0 0 0# 2 1 1 0 0# 3 0 1 1 0# 4 1 1 1 1
答案2
得分: 2
以下是您要翻译的内容：
Base R方式，通用于希望适用于任意数量不同字符串的实际数据的情况：# 更新的示例数据框 - NA不应被引用，去除尾随逗号df <- data.frame(stress1 = c('A', 'A', 'B', 'A'),stress2 = c(NA, 'B', 'C', 'B'),stress3 = c(NA, NA, NA, 'C'),stress4 = c(NA, NA, NA, 'D'))desiredout <- data.frame (A = c(1, 1, 0, 1),B = c(0, 1, 1, 1),C = c(0, 0, 1, 1),D = c(0, 0, 0, 1))out = data.frame( # 用于数据框输出lapply( # 迭代unique(unlist(df[!is.na(df)]) # df中的所有唯一非NA值),\(x) {rowSums(df == x, na.rm = TRUE)}))names(out) <- unique(unlist(df[!is.na(df)]))
检查：
> all.equal(desiredout, out)[1] TRUE
以及：
> outA B C D1 1 0 0 02 1 1 0 03 0 1 1 04 1 1 1 1
如果您发现在最后一列中除了1或0以外的数字，那么可能有行中有多个相同字符串的实例 - 如果是这样，请回来，我们可以进行相应的编辑。
英文:
Base R way, generalised to hopefully work for real data with any number of different strings in:
# updated sample data frame - NA's should not be quoted, removed trailing commadf <- data.frame(stress1 = c('A', 'A', 'B', 'A'),stress2 = c(NA, 'B', 'C', 'B'),stress3 = c(NA, NA, NA, 'C'),stress4 = c(NA, NA, NA, 'D'))desiredout <- data.frame (A = c(1, 1, 0, 1),B = c(0, 1, 1, 1),C = c(0, 0, 1, 1),D = c(0, 0, 0, 1))out = data.frame( # for data frame outputlapply( # iterativeunique(unlist(df[!is.na(df)]) # all unique, non-NA values in df),\(x) {rowSums(df == x, na.rm = TRUE)}))names(out) <- unique(unlist(df[!is.na(df)]))
check:
> all.equal(desiredout, out)[1] TRUE
and:
> outA B C D1 1 0 0 02 1 1 0 03 0 1 1 04 1 1 1 1
If you find you have figures other than 1 or 0 in your final column then you may have rows with more than one instance of a given string - if so, come back and we can edit appropriately.
答案3
得分: 2
A purrr solution with pmap_dfr + table:
library(purrr)pmap_dfr(df, ~ unclass(table(c(...)))) %>% replace(is.na(.), 0)
英文:
A purrr solution with pmap_dfr + table:
library(purrr)pmap_dfr(df, ~ unclass(table(c(...)))) %>%replace(is.na(.), 0)# # A tibble: 4 × 4# A B C D# <int> <int> <int> <int># 1 1 0 0 0# 2 1 1 0 0# 3 0 1 1 0# 4 1 1 1 1
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