如何用列中最小值的一半替换零？

Question

我试图用该列中大于零的最小值的一半替换我的数千行和列的数据框中的 0。我也不想包括前四列，因为它们是索引。

所以如果我从这样的事情开始：

index <- c("100p", "200p", 300p" 400p")
ratio <- c(5, 4, 3, 2)
gene <- c("gapdh", NA, NA,"actb"
species <- c("mouse", NA, NA, "rat")
a1 <- c(0,3,5,2)
b1 <- c(0, 0, 4, 6)
c1 <- c(1, 2, 3, 4)

as.data.frame(q) <- cbind(index, ratio, gene, species, a1, b1, c1)

index ratio gene  species a1 b1 c1
100p    5   gapdh mouse   0  0  1
200p    4    NA    NA     3  0  2
300p    3    NA    NA     5  4  3
400p    2   actb  rat     2  6  4

我希望得到这样的结果：

index ratio gene  species a1 b1 c1
100p    5   gapdh mouse   1  2  1
200p    4    NA    NA     3  2  2
300p    3    NA    NA     5  4  3
400p    2   actb  rat     2  6  4

我试过下面的代码： apply(q[-4], 2, function(x) "[<-"(x, x==0, min(x[x > 0]) / 2))

但我一直收到错误消息：Error in min(x[x > 0])/2 : non-numeric argument to binary operator

有什么帮助吗？非常感谢！

Answer 1

我们可以使用 lapply 和 replace 列中最小值的 0 个值乘以 2。

cols<- 5:7
q[cols] <- lapply(q[cols], function(x) replace(x, x == 0, min(x[x>0], na.rm = TRUE)/2))

q
#  index ratio  gene species a1 b1 c1
#1  100p     5 gapdh   mouse  1  2  1
#2  200p     4  <NA>    <NA>  3  2  2
#3  300p     3  <NA>    <NA>  5  4  3
#4  400p     2  actb     rat  2  6  4

在dplyr中，我们可以使用mutate_at

library(dplyr)
q %>%  mutate_at(cols,~replace(., . == 0, min(.[.>0], na.rm = TRUE)/2))

数据

q <- structure(list(index = structure(1:4, .Label = c("100p", "200p", 
"300p", "400p"), class = "factor"), ratio = c(5, 4, 3, 2), gene = structure(c(2L, 
NA, NA, 1L), .Label = c("actb", "gapdh"), class = "factor"), 
species = structure(c(1L, NA, NA, 2L), .Label = c("mouse", 
"rat"), class = "factor"), a1 = c(0, 3, 5, 2), b1 = c(0, 
0, 4, 6), c1 = c(1, 2, 3, 4)), class = "data.frame", row.names = c(NA, -4L))

Answer 2

稍有不同（对于大型数据集可能更快）dplyr 带有一点数学运算的选项可能是：

q %>%
 mutate_at(vars(5:length(.)), ~ (. == 0) * min(.[. != 0])/2 + .)

  index ratio  gene species a1 b1 c1
1  100p     5 gapdh   mouse  1  2  1
2  200p     4  <NA>    <NA>  3  2  2
3  300p     3  <NA>    <NA>  5  4  3
4  400p     2  actb     rat  2  6  4

与base R相同：

q[, 5:length(q)] <- lapply(q[, 5:length(q)], function(x) (x == 0) * min(x[x != 0])/2 + x)

Answer 3

供参考，考虑到您的原始代码，我相信您的功能不是问题所在。相反，错误来自将函数应用于非数字数据。

# original data
index <- c("100p", "200p", "300p" , "400p")
ratio <- c(5, 4, 3, 2)
gene <- c("gapdh", NA, NA,"actb")
species <- c("mouse", NA, NA, "rat")
a1 <- c(0,3,5,2)
b1 <- c(0, 0, 4, 6)
c1 <- c(1, 2, 3, 4)

# data frame
q <- as.data.frame(cbind(index, ratio, gene, species, a1, b1, c1))

# examine structure (all cols are factors) 
str(q)

# convert factors to numeric  
fac_to_num <- function(x){
  x <- as.numeric(as.character(x))
  x
}

# apply to cols 5 thru 7 only
q[, 5:7] <- apply(q[, 5:7],2,fac_to_num)

# examine structure  
str(q)

# use original function only on numeric data 
apply(q[, 5:7], 2, function(x) "[<-"(x, x==0, min(x[x > 0]) / 2))

如何用列中最小值的一半替换零？

How can I replace zeros with half the minimum value within a column?

replace

r

function

min

zero