矩阵中的循环效率问题

问题描述 投票:0回答:1

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位数字。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?

r dataframe loops matrix dataset
1个回答
0
投票

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
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