我试图用另外两列中的值替换两列中的值。这是一个相当基本的问题,已被by python
users询问,但是我使用的是R。
我有一个df
看起来像这样(仅在更大范围内[> 20,000]):
squirrel_id locx locy dist
6391 17.5 10.0 50.0
6391 17.5 10.0 20.0
6391 17.5 10.0 15.5
8443 20.5 1.0 800
6025 -5.0 -0.5 0.0
对于63只松鼠,我需要替换它们的locx
和locy
值。
我通常用以下代码替换值:
library(dplyr)
df <- df %>%
mutate(locx = ifelse (squirrel_id=="6391", "12.5", locx),
locy = ifelse (squirrel_id=="6391", "15.5", locy),
locx = ifelse (squirrel_id=="8443", "2.5", locx),
locy = ifelse (squirrel_id=="8443", "80", locy)) #etc for 63 squirrels
哪个会给我的?
squirrel_id locx locy dist
6391 12.5 10.0 50.0
6391 12.5 10.0 20.0
6391 12.5 10.0 15.5
8443 2.5 80.0 800
6025 -5.0 -0.5 0.0
但这会创建额外的126行代码,我怀疑有更简单的方法来实现这一目标。
我确实在单独的locx
中拥有所有新的locy
和df
值,但是我不知道如何将{{1 }},而不会弄乱数据。
dataframe
的值需要替换旧squirrel_id
中的值:
df
我如何才能更有效地做到这一点?
答案 0 :(得分:1)
您可以left_join
两个数据帧,然后使用if_else
语句获取正确的locx
和locy
。试试:
library(dplyr)
df %>% left_join(df2, by = "squirrel_id") %>%
mutate(locx = if_else(is.na(new_locx), locx, new_locx), # as suggested by @echasnovski, we can also use locx = coalesce(new_locx, locx)
locy = if_else(is.na(new_locy), locy, new_locy)) %>% # or locy = coalesce(new_locy, locy)
select(-new_locx, -new_locy)
# output
squirrel_id locx locy dist
1 6391 12.5 15.5 50.0
2 6391 12.5 15.5 20.0
3 6391 12.5 15.5 15.5
4 8443 2.5 80.0 800.0
5 6025 -55.0 0.0 0.0
6 5000 18.5 18.5 10.0 # squirrel_id 5000 was created for an example of id
# present if df but not in df2
数据
df <- structure(list(squirrel_id = c(6391L, 6391L, 6391L, 8443L, 6025L,
5000L), locx = c(17.5, 17.5, 17.5, 20.5, -5, 18.5), locy = c(10,
10, 10, 1, -0.5, 12.5), dist = c(50, 20, 15.5, 800, 0, 10)), class = "data.frame", row.names = c(NA,
-6L))
df2 <- structure(list(squirrel_id = c(6391L, 8443L, 6025L), new_locx = c(12.5,
2.5, -55), new_locy = c(15.5, 80, 0)), class = "data.frame", row.names = c(NA,
-3L))
答案 1 :(得分:0)
使用@ANG的数据,这是一个data.table
解决方案。它通过引用连接并更新原始df
。
library(data.table)
setDT(df)
setDT(df2)
df[df2, on = c('squirrel_id'), `:=` (locx = new_locx, locy = new_locy) ]
df
squirrel_id locx locy dist
1: 6391 12.5 15.5 50.0
2: 6391 12.5 15.5 20.0
3: 6391 12.5 15.5 15.5
4: 8443 2.5 80.0 800.0
5: 6025 -55.0 0.0 0.0
6: 5000 18.5 12.5 10.0
另请参阅:
how to use merge() to update a table in R