我有以下从计数中获得的数据框。我曾经dput使数据框可用,然后编辑数据框,因此存在A.
df <- structure(list(Procedure = structure(c(4L, 1L, 2L, 3L), .Label = c("A", "A", "C", "D", "-1"),
class = "factor"), n = c(10717L, 4412L, 2058L, 1480L)),
class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -4L), .Names = c("Procedure", "n"))
print(df)
# A tibble: 4 x 2
Procedure n
<fct> <int>
1 D 10717
2 A 4412
3 A 2058
4 C 1480
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现在我想对过程进行区分,只保留第一个A。
df %>%
distinct(Procedure, .keep_all=TRUE)
# A tibble: 4 x 2
Procedure n
<fct> <int>
1 D 10717
2 A 4412
3 A 2058
4 C 1480
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这是行不通的。奇怪的...
如果我们打印该Procedure列,我们可以看到 存在重复的级别a,这对于该distinct函数来说是有问题的。
df$Procedure
[1] D A A C
Levels: A A C D -1
Warning message:
In print.factor(x) : duplicated level [2] in factor
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解决方法之一是降低因子水平。我们可以使用factor函数来实现这一点。另一种方法是将Procedure列转换为字符。
df <- structure(list(Procedure = structure(c(4L, 1L, 2L, 3L), .Label = c("A", "A", "C", "D", "-1"),
class = "factor"), n = c(10717L, 4412L, 2058L, 1480L)),
class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -4L), .Names = c("Procedure", "n"))
library(tidyverse)
df %>%
mutate(Procedure = factor(Procedure)) %>%
distinct(Procedure, .keep_all=TRUE)
# # A tibble: 3 x 2
# Procedure n
# <fct> <int>
# 1 D 10717
# 2 A 4412
# 3 C 1480
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