我有一个640 x 2500的数据框，里面有数值和几个 NA 值。我的目标是找到至少75个连续的 NA 的值。对于每一个这样的运行，我想替换之前的 和 以下50个细胞 NA 值。
下面是一个缩小的例子，是一行。
x <- c(1, 3, 4, 5, 4, 3, NA, NA, NA, NA, 6, 9, 3, 2, 4, 3)
# run of four NA: ^ ^ ^ ^
我想检测连续四次的运行 NA，然后将运行前的三个值和运行后的三个值替换为 NA:
c(1, 3, 4, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 2, 4, 3)
# ^ ^ ^ ^ ^ ^
我曾试图首先确定连续 NA与 rle但是运行 rle(is.na(df)) 给出错误 'x' must be a vector of an atomic type. 即使在我选择单行时也会出现这种情况。
遗憾的是，我不知道下一步该如何将之前和之后的50个单元格转换为NA。
如果能得到任何帮助，我将不胜感激，提前表示感谢。
因为你评论说，在你的数据中"有的[行]以数行开头，有的[行]以数行结尾。NAs"，希望这能更好地代表真实数据。
A B C D E F G H I J
1 1 2 3 NA NA 6 7 8 NA 10
2 1 NA NA NA 5 6 7 NA NA NA
3 1 2 3 4 NA NA NA 8 9 10
让我们假设最小运行长度为： NA 拟扩大为 NA 是2，运行前的两个值和运行后的两个值应替换为 NA. 在这个例子中，第2行将代表你在评论中提到的情况。
首先是一些数据的处理。我更喜欢用 data.table 在 长 格式。随着 data.table 我们可以获得有用的常数 .I 和 .N，并且可以轻松地用 rleid.
# convert data.frame to data.table
library(data.table)
setDT(d)
# set minimum length of runs to be expanded
len = 2L
# set number of values to replace on each side of run
n = 2L
# number of columns of original data (for truncation of indices)
nc = ncol(d)
# create a row index to keep track of the original rows in the long format
d[ , ri := 1:.N]
# melt from wide to long format
d2 = melt(d, id.vars = "ri")
# order by row index
setorder(d2, ri)
现在，实际计算的运行和他们的指数。
d2[
# check if the run is an "NA run" and has sufficient length
d2[ , if(anyNA(value) & .N >= len){
# get indices before and after run, where values should be changed to NA
ix = c(.I[1] - n:1L, .I[.N] + 1L:n)
# truncate indices to keep them within (original) rows
ix[ix >= 1 + (ri - 1) * nc & ix <= nc * ri]},
# perform the calculation by row index and run index
# grab replacement indices
by = .(ri, rleid(is.na(value)))]$V1,
# at replacement indices, set value to NA
value := NA]
如果需要的话，可以投回宽幅
dcast(d2, ri ~ variable, value.vars = "value")
# ri A B C D E F G H I J
# 1: 1 1 NA NA NA NA NA NA 8 NA 10
# 2: 2 NA NA NA NA NA NA NA NA NA NA
# 3: 3 1 2 NA NA NA NA NA NA NA 10
类型胁迫对我有效。
rle(as.logical(is.na(x[MyRow, ])))
这是我的解决方案。不过不知道有没有比我的更简洁的解决方案。
library(data.table)
df <- matrix(nrow = 1,ncol = 16)
df[1,] <- c(1, 3, 4, 5, 4, 3, NA, NA, NA, NA, 6, 9, 3, 2, 4, 3)
df <- df %>%
as.data.table() # dataset created
# A function to do what you need
NA_replacer <- function(x){
Vector <- unlist(x) # pull the values into a vector
NAs <- which(is.na(Vector)) # locate the positions of the NAs
NAs_Position_1 <- cumsum(c(1, diff(NAs) - 1)) # Find those that are in sequential order
NAs_Position_2 <- rle(NAs_Position_1) # Find their values
NAs <- NAs[which(
NAs_Position_1 == with(NAs_Position_2,
values[which(
lengths == 4)]))] # Locate the position of those NAs that are repeated exactly 4 times
if(length(NAs == 4)){ # Check if there are a stretch of 4 WAs
Vector[seq(NAs[1]-3,
NAs[1]-1,1)] <- NA # this part deals with the 3 positions occuring before the first NA
Vector[seq(NAs[length(NAs)]+1,
NAs[length(NAs)]+3,1)] <- NA # this part deals with the 3 positions occuring after the last NA
}
Vector
}
> df # the original dataset
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16
1: 1 3 4 5 4 3 NA NA NA NA 6 9 3 2 4 3
# the transformed dataset
apply(df, 1, function(x) NA_replacer(x)) %>%
as.data.table() %>%
data.table::transpose()
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16
1: 1 3 4 NA NA NA NA NA NA NA NA NA NA 2 4 3
说句题外话，对于一个大小为640*2500的假设数据框来说，速度是相当不错的，其中有一段75个或更多的NA必须被定位，而且前后50个值必须用一个NA来代替。
df <- matrix(nrow = 640,ncol = 2500)
for(i in 1:nrow(df)){
df[i,] <- c(1:100,rep(NA,75),rep(1,2325))
}
NA_replacer <- function(x){
Vector <- unlist(x) # pull the values into a vector
NAs <- which(is.na(Vector)) # locate the positions of the NAs
NAs_Position_1 <- cumsum(c(1, diff(NAs) - 1)) # Find those that are in sequential order
NAs_Position_2 <- rle(NAs_Position_1) # Find their values
NAs <- NAs[which(
NAs_Position_1 == with(NAs_Position_2,
values[which(
lengths >= 75)]))] # Locate the position of those NAs that are repeated exactly 75 times or more than 75 times
if(length(NAs >= 75)){ # Check if the condition is met
Vector[seq(NAs[1]-50,
NAs[1]-1,1)] <- NA # this part deals with the 50 positions occuring before the first NA
Vector[seq(NAs[length(NAs)]+1,
NAs[length(NAs)]+50,1)] <- NA # this part deals with the 50 positions occuring after the last NA
}
Vector
}
# Check how many NAs are present in the first row of the dataset prior to applying the function
which(is.na(df %>%
as_tibble() %>%
slice(1) %>%
unlist())) %>% # run the code till here to get the indices of the NAs
length()
[1] 75
df <- apply(df, 1, function(x) NA_replacer(x)) %>%
as.data.table() %>%
data.table::transpose()
# Check how many NAs are present in the first row post applying the function
which(is.na(df %>%
slice(1) %>%
unlist())) %>% # run the code till here to get the indices of the NAs
length()
[1] 175
system.time(df <- apply(df, 1, function(x) NA_replacer(x)) %>%
as.data.table() %>%
data.table::transpose())
user system elapsed
0.216 0.002 0.220