Question

我有这个数据框

    # of int.   int.   not.int.  ID  group   odd      even
2      24      85.15    113.34   2   thc1    NA       486.66
3      33      134.94   158.17   3   thc2    465.06   NA
4      12      47.60    62.73    4   thc3    NA       537.27
1      50      218.41   372.16   1   veh     381.59   NA
5      44      176.81   268.92   5   veh     NA       331.08

如何替换“ int”中的数字。带有“偶数”的数字而不添加NA，以及“奇数”和“ not.int”的数字相同？所以看起来像这样。

 # of int.   int.     not.int.  ID  group   odd      even
2      24    486.66   113.34    2   thc1    NA       486.66
3      33    134.94   465.06    3   thc2    465.06   NA
4      12    537.27   62.73     4   thc3    NA       537.27
1      50    218.41   381.59    1   veh     381.59   NA
5      44    331.08   268.92    5   veh     NA       331.08

Answer 1

一种选择是使用Map中的base R来获取相应的列，然后进行赋值

df1[2:3] <- Map(function(x, y) {
              i1 <- !is.na(y)
              x[i1] <- y[i1]
               x}, df1[c('int.', 'not.int.')], df1[c('even', 'odd')])

df1
#  # of int.   int. not.int. ID group    odd   even
#2        24 486.66   113.34  2  thc1     NA 486.66
#3        33 134.94   465.06  3  thc2 465.06     NA
#4        12 537.27    62.73  4  thc3     NA 537.27
#1        50 218.41   381.59  1   veh 381.59     NA
#5        44 331.08   268.92  5   veh     NA 331.08

数据

 df1 <- structure(list(`# of int.` = c(24L, 33L, 12L, 50L, 44L), int. = c(85.15, 
134.94, 47.6, 218.41, 176.81), not.int. = c(113.34, 158.17, 62.73, 
372.16, 268.92), ID = c(2L, 3L, 4L, 1L, 5L), group = c("thc1", 
"thc2", "thc3", "veh", "veh"), odd = c(NA, 465.06, NA, 381.59, 
NA), even = c(486.66, NA, 537.27, NA, 331.08)), .Names = c("# of int.", 
"int.", "not.int.", "ID", "group", "odd", "even"), 
class = "data.frame", row.names = c("2", 
"3", "4", "1", "5"))

Answer 2

使用mutate包中的ifelse和dplyr：

library(dplyr)

df %>%
 mutate(int. = ifelse(is.na(even), int., even),
        not.int. = ifelse(is.na(odd), not.int., odd))

Answer 3

这是基数R中的一个选项，带有ifelse和is.na。

dat$int <- with(dat, ifelse(!is.na(even), even, int))
dat$not.int <- with(dat, ifelse(!is.na(odd), odd, not.int))

数据

dat <- read.table(text = "    '# of int'   int   'not.int'  ID  group   odd      even
2      24      85.15    113.34   2   thc1    NA       486.66
3      33      134.94   158.17   3   thc2    465.06   NA
4      12      47.60    62.73    4   thc3    NA       537.27
1      50      218.41   372.16   1   veh     381.59   NA
5      44      176.81   268.92   5   veh     NA       331.08",
                  header = TRUE, stringsAsFactors = FALSE)

Answer 4

x <- df1[c("even","odd")]
df1[c("int.","not.int.")][!is.na(x)] <- x[!is.na(x)]

# #   of int.   int. not.int. ID group    odd   even
# 2        24 486.66   113.34  2  thc1     NA 486.66
# 3        33 134.94   465.06  3  thc2 465.06     NA
# 4        12 537.27    62.73  4  thc3     NA 537.27
# 1        50 218.41   381.59  1   veh 381.59     NA
# 5        44 331.08   268.92  5   veh     NA 331.08

R中不带NA的行替换

4 个答案:

数据