将行添加到数据框

Adding row to a dataframe

我有一个这样的 df:

x <- data.frame("Year" = c(1945,1945,1946,1946,1947,1947), "Age" = c(1,2,1,2,1,2), "Value" = c(4,5,4,5,4,6))

我想根据某些条件添加新行: 例如,我想添加与第一行具有相同“YEAR”和“AGE”的第二行,但值是第一行的值减去 1。

我想要这样的结果:

x <- data.frame("Year" = c(1945, 1945,1945,1945,1946,1946,1946,1946,1947,1947,1947,1947), "Age" = c(1,1,2,2,1,1,2,2,1,1,2,2), "Value" = c(4,3,5,4,4,3,5,4,4,3,5,4))

感谢您帮助初学者。

你可以这样做:

#original df
x <- data.frame("Year" = c(1945,1945,1946,1946,1947,1947), "Age" = c(1,2,1,2,1,2),"Value" = c(4,5,4,5,4,6))

#replicate df with value minus 1
y <- data.frame(x[,c("Year", "Age")], Value = x[,"Value"] -1)

#combine
z <- rbind(x,y)
z[order(z$Year, z$Age),]

   Year Age Value
1  1945   1     4
7  1945   1     3
2  1945   2     5
8  1945   2     4
3  1946   1     4
9  1946   1     3
4  1946   2     5
10 1946   2     4
5  1947   1     4
11 1947   1     3
6  1947   2     6
12 1947   2     5

上面的代码当然可以缩短,例如直接更新数据框 x 时的代码。

x <- rbind(x, data.frame(x[,c("Year", "Age")], Value = x[,"Value"] -1))

正如下面的评论所指出的,可以使用 transform()within().

进一步缩短代码(并使其更具可读性)
x <- rbind(x, transform(x, Value = Value - 1))
x <- rbind(x, within(x, Value <- Value - 1))

这个有用吗:

library(dplyr)
x %>% inner_join(x %>% select(1)) %>% group_by(Year, Age) %>% 
mutate(Value = case_when(row_number() == 2 ~ Value - 1, TRUE ~ Value))
Joining, by = "Year"
# A tibble: 12 x 3
# Groups:   Year, Age [6]
    Year   Age Value
   <dbl> <dbl> <dbl>
 1  1945     1     4
 2  1945     1     3
 3  1945     2     5
 4  1945     2     4
 5  1946     1     4
 6  1946     1     3
 7  1946     2     5
 8  1946     2     4
 9  1947     1     4
10  1947     1     3
11  1947     2     6
12  1947     2     5

使用的数据:

x
  Year Age Value
1 1945   1     4
2 1945   2     5
3 1946   1     4
4 1946   2     5
5 1947   1     4
6 1947   2     6

base中,你可以将每一行重复两次,并将回收的0:-1添加到Value

within(x[rep(1:nrow(x), each = 2), ], Value <- Value + 0:-1)

它的tidyverse版本是

library(tidyverse)

x %>%
  uncount(2) %>% 
  mutate(Value = Value + 0:-1)

他们都给出了以下内容。输出已经排序好,所以你不需要使用 order()arrange().

   Year Age Value
1  1945   1     4
2  1945   1     3
3  1945   2     5
4  1945   2     4
5  1946   1     4
6  1946   1     3
7  1946   2     5
8  1946   2     4
9  1947   1     4
10 1947   1     3
11 1947   2     6
12 1947   2     5