[英]dplyr mutate() displaying NA values when matched from dataframe
I am trying to replace values found in one column of a dataframe based upon finding a match in another dataframe using mutate().我正在尝试根据使用 mutate() 在另一个 dataframe 中找到匹配项来替换在 dataframe 的一列中找到的值。 Here is an example:这是一个例子:
rename_ds <- data.frame(
car_name = c("Camaro Z28","AMC Javelin"),
replace_with = c("Camaro","Javelin"),
stringsAsFactors = FALSE)
mt_cars <- mtcars %>%
tibble::rownames_to_column() %>%
dplyr::rename("car_name" = rowname) %>%
dplyr::mutate(car_name = ifelse(car_name %in% rename_ds$car_name,
rename_ds[which(rename_ds$car_name == car_name),2],
car_name)
When I run this, instead of the car names being replaced by their respective replacements in rename_ds$replace_with, they are NA.当我运行它时,不是用 rename_ds$replace_with 中的各自替换替换汽车名称,而是它们是 NA。
21 Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
22 Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
23 <NA> 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2
24 <NA> 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4
25 Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
26 Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1
27 Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
Any suggestions?有什么建议么? Thanks in advance.提前致谢。
We could make it simpler with a named vector and coalesce
:我们可以使用命名向量和coalesce
使其更简单:
library(dplyr)
mtcars %>%
tibble::rownames_to_column("car_name") %>%
mutate(car_name = coalesce(set_names(rename_ds$replace_with,
rename_ds$car_name)[car_name], car_name))
# car_name mpg cyl disp hp drat wt qsec vs am gear carb
#1 Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
#2 Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
#3 Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
#4 Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
#5 Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
#6 Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
#7 Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
#8 Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
#9 Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
#10 Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4
#11 Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4
#12 Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
#13 Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3
#14 Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
#15 Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
#16 Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4
#17 Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4
#18 Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
#19 Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2
#20 Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1
#21 Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
#22 Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
#23 Javelin 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2
#24 Camaro 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4
#25 Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
#26 Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1
#27 Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
#28 Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
#29 Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
#30 Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
#31 Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
#32 Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2
In base R
, we could do在base R
中,我们可以做
pmax(row.names(mtcars), setNames(rename_ds$replace_with,
rename_ds$car_name)[row.names(mtcars)], na.rm = TRUE)
To me this looks more like a join operation:对我来说,这看起来更像是一个连接操作:
mtcars %>%
tibble::rownames_to_column() %>%
dplyr::rename("car_name" = rowname) %>%
left_join(rename_ds, by = "car_name") %>%
mutate(car_name = coalesce(replace_with, car_name)) %>%
select(-replace_with)
# car_name mpg cyl disp hp drat wt qsec vs am gear carb
# 1 Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
# 2 Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
# 3 Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
# 4 Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
# 5 Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
# 6 Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
# 7 Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
# 8 Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
# 9 Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
# 10 Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4
# 11 Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4
# 12 Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
# 13 Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3
# 14 Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
# 15 Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
# 16 Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4
# 17 Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4
# 18 Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
# 19 Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2
# 20 Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1
# 21 Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
# 22 Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
# 23 Javelin 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2
# 24 Camaro 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4
# 25 Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
# 26 Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1
# 27 Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
# 28 Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
# 29 Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
# 30 Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
# 31 Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
# 32 Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2
(Rows 23-24 are updated.) (第 23-24 行已更新。)
You are on the right track, you can use str_replace_all
你在正确的轨道上,你可以使用str_replace_all
mtcars %>%
tibble::rownames_to_column() %>%
dplyr::rename("car_name" = rowname) %>%
dplyr::mutate(car_name = str_replace_all(car_name,
exec(str_c,collapse="|",rename_ds$car_name),
exec(setNames,!!!unname(rev(rename_ds)))))
car_name mpg cyl disp hp drat wt qsec vs am gear carb
1 Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
2 Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
3 Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
4 Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
5 Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
6 Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
7 Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
8 Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
9 Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
10 Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4
11 Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4
12 Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
13 Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3
14 Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
15 Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
16 Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4
17 Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4
18 Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
19 Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2
20 Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1
21 Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
22 Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
23 Camaro 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 # Changed
24 Javelin 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 # Changed
25 Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
26 Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1
27 Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
28 Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
29 Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
30 Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
31 Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
32 Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.