我的数据框架有一个日期列和两个数值列，包括一些NA，如下所示:
df
# Date a b
# 1 1990-02-01 NA NA
# 2 1990-03-01 NA NA
# 3 1990-04-01 NA 3
# 4 1990-05-01 1 4
# 5 1990-06-01 2 5
# 6 1990-07-01 3 NA
# 7 1990-08-01 4 7
# 8 1990-09-01 5 NA
# 9 1990-10-01 6 9
# 10 1990-11-01 7 NA
# 11 1990-12-01 8 NA
# 12 1991-01-01 9 NA
# 13 1991-02-01 10 13
# 14 1991-03-01 11 14
# 15 1991-04-01 12 15
# 16 1991-05-01 13 NA
我想保留在时间序列开始之前出现的NAs，然后用零替换它们。最终结果应该如下所示:
finaldf
# Date a b
# 1 1990-02-01 NA NA
# 2 1990-03-01 NA NA
# 3 1990-04-01 NA 3
# 4 1990-05-01 1 4
# 5 1990-06-01 2 5
# 6 1990-07-01 3 0
# 7 1990-08-01 4 7
# 8 1990-09-01 5 0
# 9 1990-10-01 6 9
# 10 1990-11-01 7 0
# 11 1990-12-01 8 0
# 12 1991-01-01 9 0
# 13 1991-02-01 10 13
# 14 1991-03-01 11 14
# 15 1991-04-01 12 15
# 16 1991-05-01 13 0
是否有一些替换/填充功能从一些很酷的包，做到这一点?或者你自己会怎么处理这个问题?
数据df <- data.frame(Date=seq(lubridate::ymd('1990-02-01'), lubridate::ymd('1991-05-01'), by='1 month'),
a=c(rep(NA, 3), 1:13),
b=c(NA, NA, 3, 4, 5, NA, 7, NA, 9, NA, NA, NA, 13, 14, 15, NA))
finaldf <- data.frame(Date=seq(lubridate::ymd('1990-02-01'), lubridate::ymd('1991-05-01'), by='1 month'),
a=c(rep(NA, 3), 1:13),
b=c(NA, NA, 3, 4, 5, 0, 7, 0, 9, 0, 0, 0, 13, 14, 15, 0))
这很像TarJae的回答，但稍微更动态:
library(dplyr)
df %>%
mutate(across(c(a, b), ~ifelse(cumsum(!is.na(.)) > 0 & is.na(.), 0, .)))
这返回
Date a b
1 1990-02-01 NA NA
2 1990-03-01 NA NA
3 1990-04-01 NA 3
4 1990-05-01 1 4
5 1990-06-01 2 5
6 1990-07-01 3 0
7 1990-08-01 4 7
8 1990-09-01 5 0
9 1990-10-01 6 9
10 1990-11-01 7 0
11 1990-12-01 8 0
12 1991-01-01 9 0
13 1991-02-01 10 13
14 1991-03-01 11 14
15 1991-04-01 12 15
16 1991-05-01 13 0
您可以这样考虑:
df <- data.frame(Date = seq(ymd('1990-02-01'),ymd('1991-05-01'), by = '1 month'),
a = c(rep(NA,3),1:13),
b = c(NA,NA,3,4,5,NA,7,NA,9,NA,NA,NA,13,14,15,NA) )
df$b <- ifelse(is.na(df$b) & (df$Date > "1990-04-01"), 0, df$b)
df
Date a b
1 1990-02-01 NA NA
2 1990-03-01 NA NA
3 1990-04-01 NA 3
4 1990-05-01 1 4
5 1990-06-01 2 5
6 1990-07-01 3 0
7 1990-08-01 4 7
8 1990-09-01 5 0
9 1990-10-01 6 9
10 1990-11-01 7 0
11 1990-12-01 8 0
12 1991-01-01 9 0
13 1991-02-01 10 13
14 1991-03-01 11 14
15 1991-04-01 12 15
16 1991-05-01 13 0
我们可以用across检查a和b列，并结合ifelse语句:
library(dplyr)
df %>%
mutate(across(c(a, b), ~ifelse(Date > Date[4] & is.na(.), 0, .)))
#mutate(across(c(a, b), ~ifelse(Date > Date[a==1] & is.na(.), 0, .))) # more general
Date a b
<date> <int> <dbl>
1 1990-02-01 NA NA
2 1990-03-01 NA NA
3 1990-04-01 NA 3
4 1990-05-01 1 4
5 1990-06-01 2 5
6 1990-07-01 3 0
7 1990-08-01 4 7
8 1990-09-01 5 0
9 1990-10-01 6 9
10 1990-11-01 7 0
11 1990-12-01 8 0
12 1991-01-01 9 0
13 1991-02-01 10 13
14 1991-03-01 11 14
15 1991-04-01 12 15
16 1991-05-01 13 0
我们可以在which.min和which.max之间使用replace，不需要包。
u <- which.min(df$b):which.max(df$b)
df$b[u] <- replace(df$b[u], is.na(df$b[u]), 0)
df
# Date a b
# 1 1990-02-01 NA NA
# 2 1990-03-01 NA NA
# 3 1990-04-01 NA 3
# 4 1990-05-01 1 4
# 5 1990-06-01 2 5
# 6 1990-07-01 3 0
# 7 1990-08-01 4 7
# 8 1990-09-01 5 0
# 9 1990-10-01 6 9
# 10 1990-11-01 7 0
# 11 1990-12-01 8 0
# 12 1991-01-01 9 0
# 13 1991-02-01 10 13
# 14 1991-03-01 11 14
# 15 1991-04-01 12 15
# 16 1991-05-01 13 NA
数据:
df <- structure(list(Date = structure(c(7336, 7364, 7395, 7425, 7456,
7486, 7517, 7548, 7578, 7609, 7639, 7670, 7701, 7729, 7760, 7790
), class = "Date"), a = c(NA, NA, NA, 1L, 2L, 3L, 4L, 5L, 6L,
7L, 8L, 9L, 10L, 11L, 12L, 13L), b = c(NA, NA, 3, 4, 5, NA, 7,
NA, 9, NA, NA, NA, 13, 14, 15, NA)), class = "data.frame", row.names = c(NA,
-16L))
zoo::na.fill接受第二个参数，包含3个元素，用于填充前导，内导和尾NA的so:
library(zoo)
replace(df, -1, na.fill(df[-1], c(NA, 0, 0)))
或
df[-1] <- na.fill(df[-1], c(NA, 0, 0))