使用 dplyr 查找重复的元素

我尝试使用提供给 给你的代码来查找所有带有 dplyr的重复元素，如下所示:

library(dplyr)


mtcars %>%
mutate(cyl.dup = cyl[duplicated(cyl) | duplicated(cyl, from.last = TRUE)])

我怎样才能转换代码提出的 给你找到所有重复的元素与 dplyr？我上面的代码只是抛出一个错误？或者更好的是，是否还有其他功能可以比复杂的 x[duplicated(x) | duplicated(x, from.last = TRUE)])方法更简洁地实现这一点？

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最佳答案

I guess you could use filter for this purpose:

mtcars %>%
group_by(carb) %>%
filter(n()>1)

Small example (note that I added summarize() to prove that the resulting data set does not contain rows with duplicate 'carb'. I used 'carb' instead of 'cyl' because 'carb' has unique values whereas 'cyl' does not):

mtcars %>% group_by(carb) %>% summarize(n=n())
#Source: local data frame [6 x 2]
#
#  carb  n
#1    1  7
#2    2 10
#3    3  3
#4    4 10
#5    6  1
#6    8  1


mtcars %>% group_by(carb) %>% filter(n()>1) %>% summarize(n=n())
#Source: local data frame [4 x 2]
#
#  carb  n
#1    1  7
#2    2 10
#3    3  3
#4    4 10
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We can find duplicated elements with dplyr as follows. 

library(dplyr)


# Only duplicated elements
mtcars %>%
filter(duplicated(.[["carb"]])


# All duplicated elements
mtcars %>%
filter(carb %in% unique(.[["carb"]][duplicated(.[["carb"]])]))
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# Adding a shortcut to the answer above
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#>     filter, lag
#> The following objects are masked from 'package:base':
#>
#>     intersect, setdiff, setequal, union
mtcars %>% count(carb)
#> # A tibble: 6 x 2
#>    carb     n
#>   <dbl> <int>
#> 1    1.     7
#> 2    2.    10
#> 3    3.     3
#> 4    4.    10
#> 5    6.     1
#> 6    8.     1
mtcars %>% count(carb) %>% filter(n > 1)
#> # A tibble: 4 x 2
#>    carb     n
#>   <dbl> <int>
#> 1    1.     7
#> 2    2.    10
#> 3    3.     3
#> 4    4.    10


# Showing an alternative that follows the apparent intention if the asker
duplicated_carb <- mtcars %>%
mutate(dup_carb = duplicated(carb)) %>%
filter(dup_carb)
duplicated_carb
#>     mpg cyl  disp  hp drat    wt  qsec vs am gear carb dup_carb
#> 1  21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4     TRUE
#> 2  21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1     TRUE
#> 3  18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1     TRUE
#> 4  14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4     TRUE
#> 5  24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2     TRUE
#> 6  22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2     TRUE
#> 7  19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4     TRUE
#> 8  17.8   6 167.6 123 3.92 3.440 18.90  1  0    4    4     TRUE
#> 9  17.3   8 275.8 180 3.07 3.730 17.60  0  0    3    3     TRUE
#> 10 15.2   8 275.8 180 3.07 3.780 18.00  0  0    3    3     TRUE
#> 11 10.4   8 472.0 205 2.93 5.250 17.98  0  0    3    4     TRUE
#> 12 10.4   8 460.0 215 3.00 5.424 17.82  0  0    3    4     TRUE
#> 13 14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4     TRUE
#> 14 32.4   4  78.7  66 4.08 2.200 19.47  1  1    4    1     TRUE
#> 15 30.4   4  75.7  52 4.93 1.615 18.52  1  1    4    2     TRUE
#> 16 33.9   4  71.1  65 4.22 1.835 19.90  1  1    4    1     TRUE
#> 17 21.5   4 120.1  97 3.70 2.465 20.01  1  0    3    1     TRUE
#> 18 15.5   8 318.0 150 2.76 3.520 16.87  0  0    3    2     TRUE
#> 19 15.2   8 304.0 150 3.15 3.435 17.30  0  0    3    2     TRUE
#> 20 13.3   8 350.0 245 3.73 3.840 15.41  0  0    3    4     TRUE
#> 21 19.2   8 400.0 175 3.08 3.845 17.05  0  0    3    2     TRUE
#> 22 27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1     TRUE
#> 23 26.0   4 120.3  91 4.43 2.140 16.70  0  1    5    2     TRUE
#> 24 30.4   4  95.1 113 3.77 1.513 16.90  1  1    5    2     TRUE
#> 25 15.8   8 351.0 264 4.22 3.170 14.50  0  1    5    4     TRUE
#> 26 21.4   4 121.0 109 4.11 2.780 18.60  1  1    4    2     TRUE
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The original post contains an error in using the solution from the related answer. In the example given, when you use that solution inside mutate, it tries to subset the cyl vector which will not be of the same length as the mtcars dataframe.

Instead you can use the following example with filter returning all duplicated elements or mutate with ifelse to create a dummy variable which can be filtered upon later:

 library(dplyr)


# Return all duplicated elements
mtcars %>%
filter(duplicated(cyl) | duplicated(cyl, fromLast = TRUE))
# Or for making dummy variable of all duplicated
mtcars %>%
mutate(cyl.dup =ifelse(duplicated(cyl) | duplicated(cyl, fromLast = TRUE), 1,0))
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Another solution is to use janitor package:

mtcars %>% get_dupes(wt)
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Find duplicate value in data frame with column

df<-dataset[duplicated(dataset$columnname),]
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You can create a Boolean mask with duplicated():

iris %>% duplicated()
[1] FALSE FALSE FALSE .... TRUE FALSE
[145] FALSE FALSE FALSE FALSE FALSE FALSE

And pass through square brackets indexing:

iris[iris %>% duplicated(),]
Sepal.Length Sepal.Width Petal.Length Petal.Width   Species
143          5.8         2.7          5.1         1.9 virginica

Note: This approach is the closest thing to Pandas that could be done with R and dplyr:

iris[iris %>% duplicated(), c("Petal.Length","Petal.Width","Species")]
Petal.Length Petal.Width   Species
143          5.1         1.9 virginica
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A more general solution if you want to group duplicates using many columns

df%>%
select(ID,COL1,COL2,all_of(vector_of_columns))%>%
distinct%>%
ungroup%>%rowwise%>%
mutate(ID_GROUPS=paste0(ID,"_",cur_group_rows()))%>%
ungroup%>%
full_join(.,df,by=c("INFO_ID","COL1","COL2",vector_of_columns))->chk

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