R 用最新的非NA值替换NAs
在data.frame(或data.table)中,我想用最接近的前一个非NA值“前向填充”NAs。使用向量(而不是R 用最新的非NA值替换NAs,r,data.table,zoo,r-faq,R,Data.table,Zoo,R Faq,在data.frame(或data.table)中,我想用最接近的前一个非NA值“前向填充”NAs。使用向量(而不是data.frame)的简单示例如下: > y <- c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA) 我需要对许多(总容量~1 Tb)小型数据.frames(~30-50 Mb)重复此操作,其中一行是NA,其所有条目都是NA。解决问题的好方法是什么 我设计的丑陋解决方案使用以下函数: last <- function (x){
data.frame
)的简单示例如下:
> y <- c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA)
我需要对许多(总容量~1 Tb)小型数据.frame
s(~30-50 Mb)重复此操作,其中一行是NA,其所有条目都是NA。解决问题的好方法是什么
我设计的丑陋解决方案使用以下函数:
last <- function (x){
x[length(x)]
}
fill.NAs <- function(isNA){
if (isNA[1] == 1) {
isNA[1:max({which(isNA==0)[1]-1},1)] <- 0 # first is NAs
# can't be forward filled
}
isNA.neg <- isNA.pos <- isNA.diff <- diff(isNA)
isNA.pos[isNA.diff < 0] <- 0
isNA.neg[isNA.diff > 0] <- 0
which.isNA.neg <- which(as.logical(isNA.neg))
if (length(which.isNA.neg)==0) return(NULL) # generates warnings later, but works
which.isNA.pos <- which(as.logical(isNA.pos))
which.isNA <- which(as.logical(isNA))
if (length(which.isNA.neg)==length(which.isNA.pos)){
replacement <- rep(which.isNA.pos[2:length(which.isNA.neg)],
which.isNA.neg[2:max(length(which.isNA.neg)-1,2)] -
which.isNA.pos[1:max(length(which.isNA.neg)-1,1)])
replacement <- c(replacement, rep(last(which.isNA.pos), last(which.isNA) - last(which.isNA.pos)))
} else {
replacement <- rep(which.isNA.pos[1:length(which.isNA.neg)], which.isNA.neg - which.isNA.pos[1:length(which.isNA.neg)])
replacement <- c(replacement, rep(last(which.isNA.pos), last(which.isNA) - last(which.isNA.pos)))
}
replacement
}
输出
。。。这似乎有效。但是,老兄,它丑吗!有什么建议吗?您可能希望使用软件包中的
na.locf()
函数,将上一次观察结果向前推进,以替换na值
以下是帮助页面中的使用示例的开头:
library(zoo)
az <- zoo(1:6)
bz <- zoo(c(2,NA,1,4,5,2))
na.locf(bz)
1 2 3 4 5 6
2 2 1 4 5 2
na.locf(bz, fromLast = TRUE)
1 2 3 4 5 6
2 1 1 4 5 2
cz <- zoo(c(NA,9,3,2,3,2))
na.locf(cz)
2 3 4 5 6
9 3 2 3 2
图书馆(动物园)
az很抱歉,我问了一个老问题。
我在火车上找不到做这项工作的函数,所以我自己写了一个
我很自豪地发现它的速度快了一点点。
但它的灵活性较差
但它与ave
配合得很好,这正是我所需要的
repeat.before = function(x) { # repeats the last non NA value. Keeps leading NA
ind = which(!is.na(x)) # get positions of nonmissing values
if(is.na(x[1])) # if it begins with a missing, add the
ind = c(1,ind) # first position to the indices
rep(x[ind], times = diff( # repeat the values at these indices
c(ind, length(x) + 1) )) # diffing the indices + length yields how often
} # they need to be repeated
x = c(NA,NA,'a',NA,NA,NA,NA,NA,NA,NA,NA,'b','c','d',NA,NA,NA,NA,NA,'e')
xx = rep(x, 1000000)
system.time({ yzoo = na.locf(xx,na.rm=F)})
## user system elapsed
## 2.754 0.667 3.406
system.time({ yrep = repeat.before(xx)})
## user system elapsed
## 0.597 0.199 0.793
编辑
当这成为我最上等的答案时,我经常被提醒不要使用自己的函数,因为我经常需要zoo的maxgap
参数。因为当我使用无法调试的dplyr+日期时,zoo在边缘情况下有一些奇怪的问题,所以今天我回到这里来改进我的旧函数
我在这里对改进后的函数和所有其他条目进行了基准测试。对于基本功能集,tidyr::fill
是最快的,同时也不会使边缘情况失败。@BrandonBertelsen的Rcpp条目速度更快,但在输入类型方面缺乏灵活性(由于对all.equal
的误解,他错误地测试了边缘案例)
如果您需要maxgap
,我下面的函数比zoo快(并且没有日期方面的奇怪问题)
我挂了电话
新功能
repeat\u last=函数(x,forward=TRUE,maxgap=Inf,na.rm=FALSE){
如果(!向前)x=旋转(x)#如果向后移动,则反转x两次
ind=哪个(!is.na(x))#获取非缺失值的位置
if(is.na(x[1])&&!na.rm)#如果它以na开头
ind=c(1,ind)#添加第一个位置
rep_times=diff(#将指数与长度进行差分会产生多少频率
c(ind,长度(x)+1))#它们需要重复
if(maxgapmaxgap#超过maxgap
如果(有(超过)的话{#有超过的吗?
ind=排序(c(ind[超出]+1,ind))#在间隙中添加NA
重复次数=差异(c(ind,长度(x)+1))#再次差异
}
}
x=rep(x[ind],times=rep#u times)#重复这些索引处的值
如果(!正向)x=反向(x)#第二次反向
x
}
我还将该函数放在我的(仅限Github)中。尝试此函数。它不需要动物园套餐:
# last observation moved forward
# replaces all NA values with last non-NA values
na.lomf <- function(x) {
na.lomf.0 <- function(x) {
non.na.idx <- which(!is.na(x))
if (is.na(x[1L])) {
non.na.idx <- c(1L, non.na.idx)
}
rep.int(x[non.na.idx], diff(c(non.na.idx, length(x) + 1L)))
}
dim.len <- length(dim(x))
if (dim.len == 0L) {
na.lomf.0(x)
} else {
apply(x, dim.len, na.lomf.0)
}
}
处理大数据量时,为了提高效率,我们可以使用data.table包
require(data.table)
replaceNaWithLatest <- function(
dfIn,
nameColNa = names(dfIn)[1]
){
dtTest <- data.table(dfIn)
setnames(dtTest, nameColNa, "colNa")
dtTest[, segment := cumsum(!is.na(colNa))]
dtTest[, colNa := colNa[1], by = "segment"]
dtTest[, segment := NULL]
setnames(dtTest, "colNa", nameColNa)
return(dtTest)
}
require(data.table)
替换为最新的这对我很有效:
replace_na_with_last<-function(x,a=!is.na(x)){
x[which(a)[c(1,1:sum(a))][cumsum(a)+1]]
}
> replace_na_with_last(c(1,NA,NA,NA,3,4,5,NA,5,5,5,NA,NA,NA))
[1] 1 1 1 1 3 4 5 5 5 5 5 5 5 5
> replace_na_with_last(c(NA,"aa",NA,"ccc",NA))
[1] "aa" "aa" "aa" "ccc" "ccc"
library(Rcpp)
cppFunction('NumericVector na_locf_numeric(NumericVector x) {
R_xlen_t n = x.size();
for(R_xlen_t i = 0; i<n; i++) {
if(i > 0 && !R_finite(x[i]) && R_finite(x[i-1])) {
x[i] = x[i-1];
}
}
return x;
}')
把我的帽子扔进去:
library(Rcpp)
cppFunction('IntegerVector na_locf(IntegerVector x) {
int n = x.size();
for(int i = 0; i<n; i++) {
if((i > 0) && (x[i] == NA_INTEGER) & (x[i-1] != NA_INTEGER)) {
x[i] = x[i-1];
}
}
return x;
}')
以防万一:
all.equal(
na_locf(x),
replace_na_with_last(x),
na.lomf(x),
na.locf(x),
repeat.before(x)
)
[1] TRUE
更新
对于数值向量,函数有点不同:
NumericVector na_locf_numeric(NumericVector x) {
int n = x.size();
LogicalVector ina = is_na(x);
for(int i = 1; i<n; i++) {
if((ina[i] == TRUE) & (ina[i-1] != TRUE)) {
x[i] = x[i-1];
}
}
return x;
}
NumericVector na_locf_numeric(NumericVector x){
int n=x.size();
LogicalVector ina=is_na(x);
对于(int i=1;i我尝试了以下方法:
nullIdx <- as.array(which(is.na(masterData$RequiredColumn)))
masterData$RequiredColumn[nullIdx] = masterData$RequiredColumn[nullIdx-1]
nullIdx有许多软件包提供na.locf
(na
上次观察结转)功能:
xts
-xts::na.locf
zoo
-zoo::na.locf
inputets
-inputets::na.locf
spacetime
-spacetime::na.locf
还有其他软件包,其中该函数的名称不同。这对我来说很有效,尽管我不确定它是否比其他建议更有效
rollForward <- function(x){
curr <- 0
for (i in 1:length(x)){
if (is.na(x[i])){
x[i] <- curr
}
else{
curr <- x[i]
}
}
return(x)
}
前滚跟进Brandon Bertelsen的Rcpp贡献。对我来说,NumericVector版本不起作用:它只替换了第一个NA。这是因为ina
向量只在函数开始时计算一次
相反,可以采用与IntegerVector函数完全相同的方法。以下方法对我有效:
replace_na_with_last<-function(x,a=!is.na(x)){
x[which(a)[c(1,1:sum(a))][cumsum(a)+1]]
}
> replace_na_with_last(c(1,NA,NA,NA,3,4,5,NA,5,5,5,NA,NA,NA))
[1] 1 1 1 1 3 4 5 5 5 5 5 5 5 5
> replace_na_with_last(c(NA,"aa",NA,"ccc",NA))
[1] "aa" "aa" "aa" "ccc" "ccc"
library(Rcpp)
cppFunction('NumericVector na_locf_numeric(NumericVector x) {
R_xlen_t n = x.size();
for(R_xlen_t i = 0; i<n; i++) {
if(i > 0 && !R_finite(x[i]) && R_finite(x[i-1])) {
x[i] = x[i-1];
}
}
return x;
}')
库(Rcpp)
cppFunction('NumericVector na_locf_numeric(NumericVector x)){
R_xlen_t n=x.尺寸();
对于(rxlen_t i=0;i0&&!R_有限(x[i])&&R_有限(x[i-1])){
x[i]=x[i-1];
}
}
返回x;
}')
如果需要CharacterVector版本,同样的基本方法也适用:
cppFunction('CharacterVector na_locf_character(CharacterVector x) {
R_xlen_t n = x.size();
for(R_xlen_t i = 0; i<n; i++) {
if(i > 0 && x[i] == NA_STRING && x[i-1] != NA_STRING) {
x[i] = x[i-1];
}
}
return x;
}')
cppFunction('CharacterVector na\u locf\u character(CharacterVector x){
R_xlen_t n=x.尺寸();
对于(R_xlen_t i=0;i 0和x[i]==NA_字符串和x[i-1]!=NA_字符串){
x[i]=x[i-1];
}
}
返回x;
}')
a数据。表
解决方案:
dt <- data.table(y = c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA))
dt[, y_forward_fill := y[1], .(cumsum(!is.na(y)))]
dt
y y_forward_fill
1: NA NA
2: 2 2
3: 2 2
4: NA 2
5: NA 2
6: 3 3
7: NA 3
8: 4 4
9: NA 4
10: NA 4
有一个前导的NA
有点麻烦,但我发现在前导项不缺失的情况下,执行LOCF的一种非常可读(且矢量化)的方法是:
na.省略(y)[cumsum(!is.na(y))]
一般来说,可读性稍差的修改是有效的:
c(NA,NA.省略(y))[cumsum(!is.NA(y))+1]
提供所需的输出:
c(NA,2,2,2,3,3,4,4,4)
这里是@AdamO解决方案的一个修改。这个运行得更快,因为它绕过了NA.ommit
函数。这将覆盖向量y
中的NA
值(前导NA
s除外)
z您可以使用data.table
功能nafill
,可从data.table>=1.12.3
获得
library(data.table)
nafill(y, type = "locf")
# [1] NA 2 2 2 2 3 3 4 4 4
如果向量是data.table
中的一列,也可以通过引用setnafill
来更新它:
d <- data.table(x = 1:10, y)
setnafill(d, type = "locf", cols = "y")
d
# x y
# 1: 1 NA
# 2: 2 2
# 3: 3 2
# 4: 4 2
# 5: 5 2
# 6: 6 3
# 7: 7 3
# 8: 8 4
# 9: 9 4
# 10: 10 4
…您可以通过引用i来填写它们
rollForward <- function(x){
curr <- 0
for (i in 1:length(x)){
if (is.na(x[i])){
x[i] <- curr
}
else{
curr <- x[i]
}
}
return(x)
}
library(Rcpp)
cppFunction('NumericVector na_locf_numeric(NumericVector x) {
R_xlen_t n = x.size();
for(R_xlen_t i = 0; i<n; i++) {
if(i > 0 && !R_finite(x[i]) && R_finite(x[i-1])) {
x[i] = x[i-1];
}
}
return x;
}')
cppFunction('CharacterVector na_locf_character(CharacterVector x) {
R_xlen_t n = x.size();
for(R_xlen_t i = 0; i<n; i++) {
if(i > 0 && x[i] == NA_STRING && x[i-1] != NA_STRING) {
x[i] = x[i-1];
}
}
return x;
}')
dt <- data.table(y = c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA))
dt[, y_forward_fill := y[1], .(cumsum(!is.na(y)))]
dt
y y_forward_fill
1: NA NA
2: 2 2
3: 2 2
4: NA 2
5: NA 2
6: 3 3
7: NA 3
8: 4 4
9: NA 4
10: NA 4
dt <- data.table(y = c(0, 2, -2, 0, 0, 3, 0, -4, 0, 0))
dt[, y_forward_fill := y[1], .(cumsum(y != 0))]
dt
y y_forward_fill
1: 0 0
2: 2 2
3: -2 -2
4: 0 -2
5: 0 -2
6: 3 3
7: 0 3
8: -4 -4
9: 0 -4
10: 0 -4
dt <- data.table(group = sample(c('a', 'b'), 20, replace = TRUE), y = sample(c(1:4, rep(NA, 4)), 20 , replace = TRUE))
dt <- dt[order(group)]
dt[, y_forward_fill := y[1], .(group, cumsum(!is.na(y)))]
dt
group y y_forward_fill
1: a NA NA
2: a NA NA
3: a NA NA
4: a 2 2
5: a NA 2
6: a 1 1
7: a NA 1
8: a 3 3
9: a NA 3
10: a NA 3
11: a 4 4
12: a NA 4
13: a 1 1
14: a 4 4
15: a NA 4
16: a 3 3
17: b 4 4
18: b NA 4
19: b NA 4
20: b 2 2
z <- !is.na(y) # indicates the positions of y whose values we do not want to overwrite
z <- z | !cumsum(z) # for leading NA's in y, z will be TRUE, otherwise it will be FALSE where y has a NA and TRUE where y does not have a NA
y <- y[z][cumsum(z)]
library(data.table)
nafill(y, type = "locf")
# [1] NA 2 2 2 2 3 3 4 4 4
d <- data.table(x = 1:10, y)
setnafill(d, type = "locf", cols = "y")
d
# x y
# 1: 1 NA
# 2: 2 2
# 3: 3 2
# 4: 4 2
# 5: 5 2
# 6: 6 3
# 7: 7 3
# 8: 8 4
# 9: 9 4
# 10: 10 4
d <- data.table(x = c(1, NA, 2), y = c(2, 3, NA), z = c(4, NA, 5))
# x y z
# 1: 1 2 4
# 2: NA 3 NA
# 3: 2 NA 5
setnafill(d, type = "locf")
d
# x y z
# 1: 1 2 4
# 2: 1 3 4
# 3: 2 3 5
fill.NAs <- function(x) {is_na<-is.na(x); x[Reduce(function(i,j) if (is_na[j]) i else j, seq_len(length(x)), accumulate=T)]}
fill.NAs(c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA))
[1] NA 2 2 2 2 3 3 4 4 4
replace_na_with_previous<-function (vector) {
if (is.na(vector[1]))
vector[1] <- na.omit(vector)[1]
for (i in 1:length(vector)) {
if ((i - 1) > 0) {
if (is.na(vector[i]))
vector[i] <- vector[i - 1]
}
}
return(vector)
}
df[]<-lapply(df,replace_na_with_previous)
y = c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA)
# first, transform it into a data.frame
y = as.data.frame(y)
y
1 NA
2 2
3 2
4 NA
5 NA
6 3
7 NA
8 4
9 NA
10 NA
fill(y, y, .direction = 'down')
y
1 NA
2 2
3 2
4 2
5 2
6 3
7 3
8 4
9 4
10 4