R 按位置和条件筛选器访问列
我有以下数据框:R 按位置和条件筛选器访问列,r,dplyr,symbols,R,Dplyr,Symbols,我有以下数据框: df = structure(list(age = c("F", "F", "M", "M", "M", "F", "M", "M", "F", "F", "M", "M", "F", "F", &q
df = structure(list(age = c("F", "F", "M", "M", "M", "F", "M", "M",
"F", "F", "M", "M", "F", "F", "F", "F", "F", "M", "M", "F", "F"
), gender = c(52.8547945205479, 70.617475870193, 47.6986301369863,
85.4876712328767, 56.0288204261033, 27.0219178082192, 40.8583963494959,
24.6553462722298, 80.4027397260274, 55.6684931506849, 70.6904109589041,
64.5095890410959, 45.5397260273973, 78.5909038861022, 42.4219178082192,
44.0712328767123, 77.7068493150685, 70.5199279905761, 43.7178082191781,
77.7205479452055, 74.972602739726)), row.names = c(NA, -21L), class = c("tbl_df",
"tbl", "data.frame"))
我想按性别筛选大于该性别平均年龄的年龄。
但我想通过列号而不是名称来实现这一点
所以我试着:
df %>% group_by_at(1) %>% filter_at(vars(2) > mean(vars(2))
但这不起作用
有什么建议吗?当它在一列上按分组时,它会将其从过滤器中排除。试试这个
df %>% group_by(across(1)) %>%
filter(across(1, ~ mean(.) <= .))
# A tibble: 11 x 2
# Groups: age [2]
age gender
<chr> <dbl>
1 F 70.6
2 M 85.5
3 M 56.0
4 F 80.4
5 M 70.7
6 M 64.5
7 F 78.6
8 F 77.7
9 M 70.5
10 F 77.7
11 F 75.0
df%>%分组依据(跨(1))%>%
筛选(跨(1)~平均值(%)%group_按(跨(1))%>%
过滤(1,~mean(.)位于/\u的\u所有变量在dplyr
中都已被替代。下面是一种方法
library(dplyr)
df %>%
group_by(across(1)) %>%
filter(cur_data()[[1]] > mean(cur_data()[[1]])) %>%
ungroup
# age gender
# <chr> <dbl>
# 1 F 70.6
# 2 M 85.5
# 3 M 56.0
# 4 F 80.4
# 5 M 70.7
# 6 M 64.5
# 7 F 78.6
# 8 F 77.7
# 9 M 70.5
#10 F 77.7
#11 F 75.0
库(dplyr)
df%>%
分组依据(跨(1))%>%
过滤器(cur_data()[[1]]>mean(cur_data()[[1]]))%>%
解组
#年龄性别
#
#1 F 70.6
#2米85.5
#3米56.0
#4 F 80.4
#5米70.7
#6米64.5
#7 F 78.6
#8 F 77.7
#9米70.5
#10 F 77.7
#11 F 75.0
在filter
中,使用了[[1]]
,因为分组列不包括在cur_data()
中,但是如果我有多个列,并且只想按年龄列进行过滤,会发生什么情况呢?然后”这还不够…请参阅编辑后的答案。实际上,列索引将按group_by语句中的数字减少
library(dplyr)
df %>%
group_by(across(1)) %>%
filter(cur_data()[[1]] > mean(cur_data()[[1]])) %>%
ungroup
# age gender
# <chr> <dbl>
# 1 F 70.6
# 2 M 85.5
# 3 M 56.0
# 4 F 80.4
# 5 M 70.7
# 6 M 64.5
# 7 F 78.6
# 8 F 77.7
# 9 M 70.5
#10 F 77.7
#11 F 75.0