如何根据 R 中该列中特定成员的值收集列的成员
How collect members of a column based on the value of a specific member in that column in R
在下面的数据框中,我想收集 B1 的成员,它们在 B2 中的值等于或大于 B2 中“b”的值。然后在这个新信息之后,统计每个B1成员出现了多少次。
数据框:
ID B1 B2
z1 a 2.5
z1 b 1.7
z1 c 170
z1 c 9
z1 d 3
y2 a 0
y2 b 21
y2 c 15
y2 c 101
y2 d 30
y2 d 3
y2 d 15.5
x3 a 30.8
x3 a 54
x3 a 0
x3 b 30.8
x3 c 30.8
x3 d 7
所以结果是:
ID B1 B2
z1 a 2.5
z1 c 170
z1 c 9
z1 d 3
y2 c 101
y2 d 30
x3 a 30.8
x3 a 54
x3 c 30.8
和
ID B1 count
z1 a 1
z1 c 2
z1 d 1
y2 a 0
y2 c 1
y2 d 1
x3 a 2
x3 c 1
x3 d 0
使用 tidyverse:
library(tidyverse)
df %>%
group_by(ID) %>%
filter(B2 > B2[B1 == "b"]) %>%
group_by(ID, B1) %>%
count(name = "count") %>%
as.data.frame()
#> ID B1 count
#> 1 x3 a 1
#> 2 y2 c 1
#> 3 y2 d 1
#> 4 z1 a 1
#> 5 z1 c 2
#> 6 z1 d 1
由 reprex package (v2.0.1)
于 2022-04-26 创建
按 'ID'、filter
分组,其中 'B2' 大于或等于 'B2',其中 'B1' 也为 'b'创建另一个条件 'B1' 不等于 'b'
library(dplyr)
out1 <- df1 %>%
group_by(ID) %>%
filter(any(B1 == "b") & B2 >= min(B2[B1 == "b"]), B1 != 'b')
-输出
> out1
# A tibble: 9 × 3
# Groups: ID [3]
ID B1 B2
<chr> <chr> <dbl>
1 z1 a 2.5
2 z1 c 170
3 z1 c 9
4 z1 d 3
5 y2 c 101
6 y2 d 30
7 x3 a 30.8
8 x3 a 54
9 x3 c 30.8
第二个输出将使用 summarise
进行分组以获取行数,然后使用 complete
填充缺失的组合
library(tidyr)
out1 %>%
group_by(B1, .add = TRUE) %>%
summarise(count = n(), .groups = "drop_last") %>%
complete(B1 = unique(.$B1), fill = list(count = 0)) %>%
ungroup
# A tibble: 9 × 3
ID B1 count
<chr> <chr> <int>
1 x3 a 2
2 x3 c 1
3 x3 d 0
4 y2 a 0
5 y2 c 1
6 y2 d 1
7 z1 a 1
8 z1 c 2
9 z1 d 1
数据
df1 <- structure(list(ID = c("z1", "z1", "z1", "z1", "z1", "y2", "y2",
"y2", "y2", "y2", "y2", "y2", "x3", "x3", "x3", "x3", "x3", "x3"
), B1 = c("a", "b", "c", "c", "d", "a", "b", "c", "c", "d", "d",
"d", "a", "a", "a", "b", "c", "d"), B2 = c(2.5, 1.7, 170, 9,
3, 0, 21, 15, 101, 30, 3, 15.5, 30.8, 54, 0, 30.8, 30.8, 7)),
class = "data.frame", row.names = c(NA,
-18L))
在下面的数据框中,我想收集 B1 的成员,它们在 B2 中的值等于或大于 B2 中“b”的值。然后在这个新信息之后,统计每个B1成员出现了多少次。
数据框:
ID B1 B2
z1 a 2.5
z1 b 1.7
z1 c 170
z1 c 9
z1 d 3
y2 a 0
y2 b 21
y2 c 15
y2 c 101
y2 d 30
y2 d 3
y2 d 15.5
x3 a 30.8
x3 a 54
x3 a 0
x3 b 30.8
x3 c 30.8
x3 d 7
所以结果是:
ID B1 B2
z1 a 2.5
z1 c 170
z1 c 9
z1 d 3
y2 c 101
y2 d 30
x3 a 30.8
x3 a 54
x3 c 30.8
和
ID B1 count
z1 a 1
z1 c 2
z1 d 1
y2 a 0
y2 c 1
y2 d 1
x3 a 2
x3 c 1
x3 d 0
使用 tidyverse:
library(tidyverse)
df %>%
group_by(ID) %>%
filter(B2 > B2[B1 == "b"]) %>%
group_by(ID, B1) %>%
count(name = "count") %>%
as.data.frame()
#> ID B1 count
#> 1 x3 a 1
#> 2 y2 c 1
#> 3 y2 d 1
#> 4 z1 a 1
#> 5 z1 c 2
#> 6 z1 d 1
由 reprex package (v2.0.1)
于 2022-04-26 创建按 'ID'、filter
分组,其中 'B2' 大于或等于 'B2',其中 'B1' 也为 'b'创建另一个条件 'B1' 不等于 'b'
library(dplyr)
out1 <- df1 %>%
group_by(ID) %>%
filter(any(B1 == "b") & B2 >= min(B2[B1 == "b"]), B1 != 'b')
-输出
> out1
# A tibble: 9 × 3
# Groups: ID [3]
ID B1 B2
<chr> <chr> <dbl>
1 z1 a 2.5
2 z1 c 170
3 z1 c 9
4 z1 d 3
5 y2 c 101
6 y2 d 30
7 x3 a 30.8
8 x3 a 54
9 x3 c 30.8
第二个输出将使用 summarise
进行分组以获取行数,然后使用 complete
library(tidyr)
out1 %>%
group_by(B1, .add = TRUE) %>%
summarise(count = n(), .groups = "drop_last") %>%
complete(B1 = unique(.$B1), fill = list(count = 0)) %>%
ungroup
# A tibble: 9 × 3
ID B1 count
<chr> <chr> <int>
1 x3 a 2
2 x3 c 1
3 x3 d 0
4 y2 a 0
5 y2 c 1
6 y2 d 1
7 z1 a 1
8 z1 c 2
9 z1 d 1
数据
df1 <- structure(list(ID = c("z1", "z1", "z1", "z1", "z1", "y2", "y2",
"y2", "y2", "y2", "y2", "y2", "x3", "x3", "x3", "x3", "x3", "x3"
), B1 = c("a", "b", "c", "c", "d", "a", "b", "c", "c", "d", "d",
"d", "a", "a", "a", "b", "c", "d"), B2 = c(2.5, 1.7, 170, 9,
3, 0, 21, 15, 101, 30, 3, 15.5, 30.8, 54, 0, 30.8, 30.8, 7)),
class = "data.frame", row.names = c(NA,
-18L))