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用行替换缺少的值意味着如果每行正好有n个缺少的值

matrix r

tmfmnk · 技术社区 · 7 年前

我有一个数据矩阵,每行有不同数量的缺失值。我想要的是用row替换缺少的值,这意味着如果每行缺少的值的数目为n(假设为1)。

我已经为这个问题创建了一个解决方案,但这是一个非常不雅的问题,所以我正在寻找其他的解决方案。

我的解决方案:

#SAMPLE DATA

a <- c(rep(c(1:4, NA), 2))
b <- c(rep(c(1:3, NA, 5), 2))
c <- c(rep(c(1:3, NA, 5), 2))

df <- as.matrix(cbind(a,b,c), ncol = 3, nrow = 10)

#CALCULATING THE NUMBER OF MISSING VALUES PER ROW

miss_row <- rowSums(apply(as.matrix(df), c(1,2), function(x) {
  sum(is.na(x)) +
  sum(x == "", na.rm=TRUE)
}) )

df <- cbind(df, miss_row)

#CALCULATING THE ROW MEANS FOR ROWS WITH 1 MISSING VALUE

row_mean <- ifelse(df[,4] == 1, rowMeans(df[,1:3], na.rm = TRUE), NA)

df <- cbind(df, row_mean)

2 回复 | 直到 7 年前

Cath 7 年前

以下是我在评论中提到的方式,包括更多细节:

# create your matrix
df <- cbind(a, b, c) # already a matrix, you don't need as.matrix there

# Get number of missing values per row (is.na is vectorised so you can apply it directly on the entire matrix)
nb_NA_row <- rowSums(is.na(df))

# Replace missing values row-wise by the row mean when there is N NA in the row
N <- 1 # the given example
df[nb_NA_row==N] <- rowMeans(df, na.rm=TRUE)[nb_NA_row==N]

# check df

df
#      a  b  c
# [1,] 1  1  1
# [2,] 2  2  2
# [3,] 3  3  3
# [4,] 4 NA NA
# [5,] 5  5  5
# [6,] 1  1  1
# [7,] 2  2  2
# [8,] 3  3  3
# [9,] 4 NA NA
#[10,] 5  5  5

moodymudskipper 7 年前

df <- data.frame(df)
df$miss_row <- rowSums(is.na(df))
df$row_mean <- NA
df$row_mean[df$miss_row == 1] <- rowMeans(df[df$miss_row == 1,1:3],na.rm = TRUE)
#     a  b  c miss_row row_mean
# 1   1  1  1        0       NA
# 2   2  2  2        0       NA
# 3   3  3  3        0       NA
# 4   4 NA NA        2       NA
# 5  NA  5  5        1        5
# 6   1  1  1        0       NA
# 7   2  2  2        0       NA
# 8   3  3  3        0       NA
# 9   4 NA NA        2       NA
# 10 NA  5  5        1        5

(这给出了您期望的输出,它似乎与您的文本不完全一致,但对此,请参阅注释和重复链接)

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