Question

假设我有一个数据框，其中三个列都为

> df
A    B    C
1232    27.3    0.42
1232    27.3    0.36
1232    13.1    0.15
7564    13.1    0.09
7564    13.1    0.63

所需的输出是：

        [1232]    [7564]
[13.1]   0.15    0.36
[27.3]   0.39    0

我需要在A和B中创建一个具有唯一值的矩阵作为行和列。通过将原始数据帧替换为A和B的特定值并计算C列的平均值，可以计算出矩阵中任何单元格的值。

我的代码是：

mat <- matrix(rep(0), length(unique(df$A)), nrow = length(sort(unique(df$B))))
# sort is to avoid NA
colnames(mat) <- unique(df$A)
rownames(mat) <- unique(df$B)

for (row in rownames(mat)) {
    for (col in colnames(mat)) {
        x <- subset(df, A == col & B == row)
        mat[row, col] = mean(df$C)
     }
}

考虑到我必须处理具有数千行和列的矩阵，这非常慢。如何使运行速度更快？

Answer 1

您可以结合使用aggregate()和xtabs()：

df <- read.table(header=TRUE, stringsAsFactors = FALSE, text=
"A    B    C
1232    27.3    0.42
1232    27.3    0.36
1232    13.1    0.15
7564    13.1    0.09
7564    13.1    0.63")
xtabs(C ~ B + A, data=aggregate(C ~ B + A, data=df, FUN=mean))
# > xtabs(C ~ B + A, data=aggregate(C ~ B + A, data=df, FUN=mean))
#       A
# B      1232 7564
#   13.1 0.15 0.36
#   27.3 0.39 0.00

有关其他解决方案，请阅读：How to reshape data from long to wide format?

Answer 2

Tidyverse解决方案：

library(tidyverse)

df %>% 
  group_by(A, B) %>%
  summarise(C = mean(C)) %>%
  spread(A, C)

Answer 3

您可能想要这样的东西：（使用data.table）

n <- 1e3
v <- LETTERS[1:5]
set.seed(42)
df <- data.frame(A = sample(v, n, replace = T),
                 B = sample(v, n, replace = T),
                 C = sample.int(1e2, n, replace = T))


require(data.table)
dt <- as.data.table(df)
r <- dt[, .(v = mean(C)), keyby = .(A, B)] # calculate mean for each combination
r <- dcast(r, B ~ A, value.var = 'v') # transform to your structure
rmat <- as.matrix(r[, -1]) # to matrix
rownames(rmat) <- r[[1]] # add row names
rmat[1:5, 1:5]
#          A        B        C        D        E
# A 53.00000 42.71739 53.11538 49.35000 53.14286
# B 50.62745 58.41379 60.43590 48.75000 56.56410
# C 43.75000 42.93548 55.45000 52.63415 44.27907
# D 50.00000 49.84314 57.48276 50.37143 53.16667
# E 43.95122 55.46667 55.38095 43.85366 53.22222

P.S。您发布的代码不正确。循环应该是：

for (row in rownames(mat)) {
  for (col in colnames(mat)) {
    x <- subset(df, A == col & B == row)
    mat[row, col] = mean(x$C)
  }
}

P.S.S。可以像这样优化循环：

for (row in rownames(mat)) {
  for (col in colnames(mat)) {
    i <- (df$A == col & df$B == row)
    mat[row, col] <- mean(df[i, 'C'])
  }
}

从数据框的两列中创建矩阵，并在不嵌套for循环的情况下填充到第三行

3 个答案: