# 去除na - 如何用R數據框中的零代替NA值?

## 去除na r (10)

dplyr例子：

``````library(dplyr)

df1 <- df1 %>%
mutate(myCol1 = if_else(is.na(myCol1), 0, myCol1))
``````

``````> x <- c(-1,0,1,0,NA,0,1,1)
> x1 <- replace(x,5,1)
> x1
[1] -1  0  1  0  1  0  1  1

> x1 <- replace(x,5,mean(x,na.rm=T))
> x1
[1] -1.00  0.00  1.00  0.00  0.29  0.00 1.00  1.00
``````

``````library(imputeTS)
na.replace(yourDataframe, 0)
``````

``````> x <- c(1,2,NA,NA,1,1)
> x1 <- replace(x,is.na(x),0)
> x1
[1] 1 2 0 0 1 1
``````

``````df = data.frame(col = c(1,2,NA,NA,1,1))
df <- df %>%
mutate(col = replace(col,is.na(col),0))
``````

``````n <- length(levels(data.vector))+1

data.vector <- as.numeric(data.vector)
data.vector[is.na(data.vector)] <- n
data.vector <- as.factor(data.vector)
levels(data.vector) <- c("level1","level2",...,"leveln", "NAlevel")
``````

``````  write.csv(data, "data.csv", na = "0")
``````

Datacamp提取的這個簡單函數可以幫助：

``````replace_missings <- function(x, replacement) {
is_miss <- is.na(x)
x[is_miss] <- replacement

message(sum(is_miss), " missings replaced by the value ", replacement)
x
}
``````

``````replace_missings(df, replacement = 0)
``````

``````na.zero <- function (x) {
x[is.na(x)] <- 0
return(x)
}
``````

``````na.zero(some.vector)
``````

### 聚類的其他有用的Tidyverse替代方法

Locationally：

• index `mutate_at(c(5:10), funs(replace(., is.na(.), 0)))`
• 直接引用 `mutate_at(vars(var5:var10), funs(replace(., is.na(.), 0)))`
• 固定匹配 `mutate_at(vars(contains("1")), funs(replace(., is.na(.), 0)))`
• 或代替`contains()` ，請嘗試`ends_with()``starts_with()`
• 模式匹配 `mutate_at(vars(matches("\\d{2}")), funs(replace(., is.na(.), 0)))`

（僅改變數字（列）並保留字符串（列））。

• 整數 `mutate_if(is.integer, funs(replace(., is.na(.), 0)))`
• 雙倍 `mutate_if(is.numeric, funs(replace(., is.na(.), 0)))`
• 字符串 `mutate_if(is.character, funs(replace(., is.na(.), 0)))`

## 完整的分析 -

### 測試方法：

``````# Base R:
baseR.sbst.rssgn   <- function(x) { x[is.na(x)] <- 0; x }
baseR.replace      <- function(x) { replace(x, is.na(x), 0) }
baseR.for          <- function(x) { for(j in 1:ncol(x))
x[[j]][is.na(x[[j]])] = 0 }
# tidyverse
## dplyr
library(tidyverse)
dplyr_if_else      <- function(x) { mutate_all(x, funs(if_else(is.na(.), 0, .))) }
dplyr_coalesce     <- function(x) { mutate_all(x, funs(coalesce(., 0))) }

## tidyr
tidyr_replace_na   <- function(x) { replace_na(x, as.list(setNames(rep(0, 10), as.list(c(paste0("var", 1:10)))))) }

## hybrid
hybrd.ifelse     <- function(x) { mutate_all(x, funs(ifelse(is.na(.), 0, .))) }
hybrd.rplc_all   <- function(x) { mutate_all(x, funs(replace(., is.na(.), 0))) }
hybrd.rplc_at.idx<- function(x) { mutate_at(x, c(1:10), funs(replace(., is.na(.), 0))) }
hybrd.rplc_at.nse<- function(x) { mutate_at(x, vars(var1:var10), funs(replace(., is.na(.), 0))) }
hybrd.rplc_at.stw<- function(x) { mutate_at(x, vars(starts_with("var")), funs(replace(., is.na(.), 0))) }
hybrd.rplc_at.ctn<- function(x) { mutate_at(x, vars(contains("var")), funs(replace(., is.na(.), 0))) }
hybrd.rplc_at.mtc<- function(x) { mutate_at(x, vars(matches("\\d+")), funs(replace(., is.na(.), 0))) }
hybrd.rplc_if    <- function(x) { mutate_if(x, is.numeric, funs(replace(., is.na(.), 0))) }

# data.table
library(data.table)
DT.for.set.nms   <- function(x) { for (j in names(x))
set(x,which(is.na(x[[j]])),j,0) }
DT.for.set.sqln  <- function(x) { for (j in seq_len(ncol(x)))
set(x,which(is.na(x[[j]])),j,0) }
``````

### 此分析的代碼：

``````library(microbenchmark)
# 20% NA filled dataframe of 5 Million rows and 10 columns
set.seed(42) # to recreate the exact dataframe
dfN <- as.data.frame(matrix(sample(c(NA, as.numeric(1:4)), 5e6*10, replace = TRUE),
dimnames = list(NULL, paste0("var", 1:10)),
ncol = 10))
# Running 250 trials with each replacement method
# (the functions are excecuted locally - so that the original dataframe remains unmodified in all cases)
perf_results <- microbenchmark(
hybrid.ifelse    = hybrid.ifelse(copy(dfN)),
dplyr_if_else    = dplyr_if_else(copy(dfN)),
baseR.replace    = baseR.replace(copy(dfN)),
dplyr_coalesce   = dplyr_coalesce(copy(dfN)),
hybrd.rplc_at.nse= hybrd.rplc_at.nse(copy(dfN)),
hybrd.rplc_at.stw= hybrd.rplc_at.stw(copy(dfN)),
hybrd.rplc_at.ctn= hybrd.rplc_at.ctn(copy(dfN)),
hybrd.rplc_at.mtc= hybrd.rplc_at.mtc(copy(dfN)),
hybrd.rplc_at.idx= hybrd.rplc_at.idx(copy(dfN)),
hybrd.rplc_if    = hybrd.rplc_if(copy(dfN)),
tidyr_replace_na = tidyr_replace_na(copy(dfN)),
baseR.for        = baseR.for(copy(dfN)),
DT.for.set.nms   = DT.for.set.nms(copy(dfN)),
DT.for.set.sqln  = DT.for.set.sqln(copy(dfN)),
times = 250L
)
``````

### 結果摘要

``````> perf_results
Unit: milliseconds
expr       min        lq      mean    median        uq      max neval
hybrid.ifelse 5250.5259 5620.8650 5809.1808 5759.3997 5947.7942 6732.791   250
dplyr_if_else 3209.7406 3518.0314 3653.0317 3620.2955 3746.0293 4390.888   250
baseR.sbst.rssgn 1611.9227 1878.7401 1964.6385 1942.8873 2031.5681 2485.843   250
baseR.replace 1559.1494 1874.7377 1946.2971 1920.8077 2002.4825 2516.525   250
dplyr_coalesce  949.7511 1231.5150 1279.3015 1288.3425 1345.8662 1624.186   250
hybrd.rplc_at.nse  735.9949  871.1693 1016.5910 1064.5761 1104.9590 1361.868   250
hybrd.rplc_at.stw  704.4045  887.4796 1017.9110 1063.8001 1106.7748 1338.557   250
hybrd.rplc_at.ctn  723.9838  878.6088 1017.9983 1063.0406 1110.0857 1296.024   250
hybrd.rplc_at.mtc  686.2045  885.8028 1013.8293 1061.2727 1105.7117 1269.949   250
hybrd.rplc_at.idx  696.3159  880.7800 1003.6186 1038.8271 1083.1932 1309.635   250
hybrd.rplc_if  705.9907  889.7381 1000.0113 1036.3963 1083.3728 1338.190   250
tidyr_replace_na  680.4478  973.1395  978.2678 1003.9797 1051.2624 1294.376   250
baseR.for  670.7897  965.6312  983.5775 1001.5229 1052.5946 1206.023   250
DT.for.set.nms  496.8031  569.7471  695.4339  623.1086  861.1918 1067.640   250
DT.for.set.sqln  500.9945  567.2522  671.4158  623.1454  764.9744 1033.463   250
``````

### 結果Boxplot（以對數為單位）

``````# adjust the margins to prepare for better boxplot printing
par(mar=c(8,5,1,1) + 0.1)
# generate boxplot
boxplot(opN, las = 2, xlab = "", ylab = "log(time)[milliseconds]")
``````

### 試驗的彩色編碼散點圖（以對數表示）

``````qplot(y=time/10^9, data=opN, colour=expr) +
labs(y = "log10 Scaled Elapsed Time per Trial (secs)", x = "Trial Number") +
scale_y_log10(breaks=c(1, 2, 4))
``````

### 歸因和讚賞

• Tyler Rinker和Akrun展示 Akrun 。
• alexis_laz幫助我理解使用`local()`和（與Frank的耐心幫助一起）沉默壓制在加速這些方法中扮演的角色。
• ArthurYip為了添加更新的`coalesce()`函數並更新分析。
• 格雷戈爾輕輕`data.table`函數，以便最終將它們包含在該`data.table`中。
• Base R For循環： alexis_laz
• data.table對於循環： Matt_Dowle

（當然，如果你發現這些方法有用，請盡快給予他們提升。

`tidyr`方法`replace_na`兼容的另一個`dplyr`管道兼容選項，適用於多列：

``````require(dplyr)
require(tidyr)

m <- matrix(sample(c(NA, 1:10), 100, replace = TRUE), 10)
d <- as.data.frame(m)

myList <- setNames(lapply(vector("list", ncol(d)), function(x) x <- 0), names(d))

df <- d %>% replace_na(myList)
``````

``````d\$str <- c("string", NA)

myList <- myList[sapply(d, is.numeric)]

df <- d %>% replace_na(myList)
``````