假设您有以下数据框：
df <- data.frame(industry = c("DEU_10T12", "DEU_13T15", "DEU_16", "DEU_17", "ITA_10T12", "ITA_13T15", "ITA_16", "ITA_17"),
DEU_10T12 = c(20, 24, 26, 20, 10, 0, NA, 1.5),DEU_13T15 = c(15, 16, 4.5, NA, 7.5, 5, 3, 0),
DEU_16 = c(1.5, 6, 4, 0, 0.5, 15, 3, 0.5),DEU_17 = c(NA, 20, 10, 2, 0, 0, 0, 7),
ITA_10T12 = c(0.5, 2, 3, 4, 10, 50, 2, 15), ITA_13T15 = c(25, 0, 4.5, NA, 17.5, 5, 13, 0.9),
ITA_16 = c(2, 3, 40, 20, 0.5, 15, 3, 1),ITA_17 = c(1, 9, 0.5, 2, 10, 20, 50, 7))
目标是拥有以下矩阵（它应该是数字并处理 NA 求和）：
df2 <- data.frame(industry = c("DEU_10T12", "DEU_13T15", "DEU_16", "DEU_17", "ITA_10T12", "ITA_13T15", "ITA_16", "ITA_17"),
DEU_10T12 = c(0, 0, 0, 0, 10, 0, NA, 1.5),DEU_13T15 = c(0, 0, 0, 0, 7.5, 5, 3, 0),
DEU_16 = c(0, 0, 0, 0, 0.5, 15, 3, 0.5),DEU_17 = c(0, 0, 0, 0, 0, 0, 0, 7),
ITA_10T12 = c(0.5, 2, 3, 4, 0, 0, 0, 0), ITA_13T15 = c(25, 0, 4.5, NA, 0, 0, 0, 0),
ITA_16 = c(2, 3, 40, 20, 0, 0, 0, 0),ITA_17 = c(1, 9, 0.5, 2, 0, 0, 0, 0))
新矩阵（df2，转换为数字）将镜像原始矩阵（df，也是数字）的值，除非行条目与其相应的列条目共享相同的前三个字符。在这种情况下，例如行中的 DEU_10T12 和以 DEU 开头的列，该值将设置为零，忽略任何现有的 NA 值。
我尝试如下。首先，我将 df 转换为数字，如下所示
# Extract row and column names
row_names <- df$industry
col_names <- colnames(df)[-1] # Exclude 'industry' column
# Create an empty matrix
Z <- matrix(NA, nrow = length(row_names), ncol = length(col_names), dimnames = list(row_names, col_names))
# Fill in the matrix with values from the data frame
for (i in 1:length(row_names)) {
for (j in 1:length(col_names)) {
Z[i, j] <- df[i, col_names[j]]
}
}
# Create an empty matrix for Z_narrow
Z_narrow = matrix(0, nrow = nrow(Z), ncol = ncol(Z))
# Assign row and column names
rownames(Z_narrow) = rownames(Z)
colnames(Z_narrow) = colnames(Z)
# Function to get the indices of columns to be replaced with zeros based on the first three characters of the column name
get_zero_indices <- function(col_name, row_names) {substr(col_name, 1, 3) == substr(row_names, 1, 3)}
# Loop through each row of Z to populate Z_narrow
for (i in 1:nrow(Z)) {
row_name <- rownames(Z)[i]
indices_to_zero <- sapply(colnames(Z), get_zero_indices, row_names = row_name)
Z_narrow[i, indices_to_zero] <- 0
Z_narrow[i, !indices_to_zero] <- Z[i, !indices_to_zero]
}
此代码在使用这个小数据集时可以工作，但在应用于较大的数据集时会导致 R 崩溃。有什么建议吗？
可以融化原始dataframe，如果前三个字符匹配则设置为0；然后投射回广角
library(data.table)
setDT(df)
dcast(
melt(df,id.vars = "industry")[substr(industry,1,3) == substr(variable,1,3), value:=0],
industry~variable
)
输出
industry DEU_10T12 DEU_13T15 DEU_16 DEU_17 ITA_10T12 ITA_13T15 ITA_16 ITA_17
<char> <num> <num> <num> <num> <num> <num> <num> <num>
1: DEU_10T12 0.0 0.0 0.0 0 0.5 25.0 2 1.0
2: DEU_13T15 0.0 0.0 0.0 0 2.0 0.0 3 9.0
3: DEU_16 0.0 0.0 0.0 0 3.0 4.5 40 0.5
4: DEU_17 0.0 0.0 0.0 0 4.0 NA 20 2.0
5: ITA_10T12 10.0 7.5 0.5 0 0.0 0.0 0 0.0
6: ITA_13T15 0.0 5.0 15.0 0 0.0 0.0 0 0.0
7: ITA_16 NA 3.0 3.0 0 0.0 0.0 0 0.0
8: ITA_17 1.5 0.0 0.5 7 0.0 0.0 0 0.0
另一种方法，完全不使用任何重塑：
mask = apply(df, 1, \(x) c(F,substr(x[1],1,3)==substr(names(x[2:length(x)]),1,3)))
df[t(mask)] <- 0
输出：
industry DEU_10T12 DEU_13T15 DEU_16 DEU_17 ITA_10T12 ITA_13T15 ITA_16 ITA_17
1 DEU_10T12 0.0 0.0 0.0 0 0.5 25.0 2 1.0
2 DEU_13T15 0.0 0.0 0.0 0 2.0 0.0 3 9.0
3 DEU_16 0.0 0.0 0.0 0 3.0 4.5 40 0.5
4 DEU_17 0.0 0.0 0.0 0 4.0 NA 20 2.0
5 ITA_10T12 10.0 7.5 0.5 0 0.0 0.0 0 0.0
6 ITA_13T15 0.0 5.0 15.0 0 0.0 0.0 0 0.0
7 ITA_16 NA 3.0 3.0 0 0.0 0.0 0 0.0
8 ITA_17 1.5 0.0 0.5 7 0.0 0.0 0 0.0
与@langtang相同的方法，但使用
tidyverse
library(tidyverse)
df |>
pivot_longer(-industry) |>
mutate(value = ifelse(substr(industry,1,3)==substr(name,1,3),0,value)) |>
pivot_wider()
industry DEU_10T12 DEU_13T15 DEU_16 DEU_17 ITA_10T12 ITA_13T15 ITA_16 ITA_17
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 DEU_10T12 0 0 0 0 0.5 25 2 1
2 DEU_13T15 0 0 0 0 2 0 3 9
3 DEU_16 0 0 0 0 3 4.5 40 0.5
4 DEU_17 0 0 0 0 4 NA 20 2
5 ITA_10T12 10 7.5 0.5 0 0 0 0 0
6 ITA_13T15 0 5 15 0 0 0 0 0
7 ITA_16 NA 3 3 0 0 0 0 0
8 ITA_17 1.5 0 0.5 7 0 0 0 0
在基本 R 中，不要循环遍历各个行和列，而是找到唯一的前缀并循环这些前缀：
out <- as.matrix(df[, -1])
rownames(out) <- df[, 1]
prefixes <- out |>
colnames() |>
substr(1, 3) |>
unique()
prefixes <- paste0("^", prefixes)
for (pfx in prefixes) {
out[grepl(pfx, rownames(out)),
grepl(pfx, colnames(out))] <- 0
}
结果：
#> out
DEU_10T12 DEU_13T15 DEU_16 DEU_17 ITA_10T12 ITA_13T15 ITA_16 ITA_17
DEU_10T12 0.0 0.0 0.0 0 0.5 25.0 2 1.0
DEU_13T15 0.0 0.0 0.0 0 2.0 0.0 3 9.0
DEU_16 0.0 0.0 0.0 0 3.0 4.5 40 0.5
DEU_17 0.0 0.0 0.0 0 4.0 NA 20 2.0
ITA_10T12 10.0 7.5 0.5 0 0.0 0.0 0 0.0
ITA_13T15 0.0 5.0 15.0 0 0.0 0.0 0 0.0
ITA_16 NA 3.0 3.0 0 0.0 0.0 0 0.0
ITA_17 1.5 0.0 0.5 7 0.0 0.0 0 0.0