I have data:-
我有数据: -
name row
a 1
a 2
a 3
a 4
a 5
b 1
b 2
b 3
b 4
b 5
b 6
b 7
b 8
b 9
b 10
b 11
b 12
b 13
b 14
b 15
.......
It's grouped by name.
它按名称分组。
Row is column with row_number of grouped name. I need calculate new column with value that If the value in the row column is greater than 11, start counting again.
Row是具有分组名称的row_number的列。我需要计算新列的值,如果行列中的值大于11,则再次开始计数。
Should look like this:-
应该是这样的: -
name row new_row
a 1 1
a 2 2
a 3 3
a 4 4
a 5 5
b 1 1
b 2 2
b 3 3
b 4 4
b 5 5
b 6 6
b 7 7
b 8 8
b 9 9
b 10 10
b 11 11
b 12 1
b 13 2
b 14 3
b 15 4
.............
4 个解决方案
#1
3
Try this, using dplyr:
试试这个,使用dplyr:
df <- data.frame(name=c(rep("a", 5), rep("b", 13)), stringsAsFactors = FALSE)
library(dplyr)
df %>%
group_by(name) %>%
mutate(
row = row_number(),
new_row = (row - 1L) %% 11L + 1L
) %>%
ungroup()
# # A tibble: 18 × 3
# name row new_row
# <chr> <int> <int>
# 1 a 1 1
# 2 a 2 2
# 3 a 3 3
# 4 a 4 4
# 5 a 5 5
# 6 b 1 1
# 7 b 2 2
# 8 b 3 3
# 9 b 4 4
# 10 b 5 5
# 11 b 6 6
# 12 b 7 7
# 13 b 8 8
# 14 b 9 9
# 15 b 10 10
# 16 b 11 11
# 17 b 12 1
# 18 b 13 2
Using base-R:
do.call(
rbind.data.frame,
by(df, df$name,
function(x) within(x, {
row = seq_len(nrow(x))
new_row = (row - 1L) %% 11L + 1L
})
))
# name new_row row
# a.1 a 1 1
# a.2 a 2 2
# a.3 a 3 3
# a.4 a 4 4
# a.5 a 5 5
# b.6 b 1 1
# b.7 b 2 2
# b.8 b 3 3
# b.9 b 4 4
# b.10 b 5 5
# b.11 b 6 6
# b.12 b 7 7
# b.13 b 8 8
# b.14 b 9 9
# b.15 b 10 10
# b.16 b 11 11
# b.17 b 1 12
# b.18 b 2 13
#2
2
Another idea using ave from base R,
使用基础R的ave的另一个想法,
with(df, ave(row, name, FUN = function(i) replace(i, i>11, seq(i[i > 11]))))
#[1] 1 2 3 4 5 1 2 3 4 5 6 7 8 9 10 11 1 2 3 4
Good catch on the limitation and suggestion by @r2evans,
很好地了解了@ r2evans的限制和建议,
with(df, ave(row, name, FUN = function(i) (i-1)%%11+1))
#3
0
Using the base R functions ave and rep, we can do
使用基本R函数ave和rep,我们可以做到
ave(df$row, df$name, FUN=function(x) rep(1:11, length.out=length(x)))
[1] 1 2 3 4 5 1 2 3 4 5 6 7 8 9 10 11 1 2 3 4
This uses with reps length.out argument to repeat the sequence (1:11) truncating or repeating according to the length of each group.
这与reps length.out参数一起使用,重复序列(1:11)根据每个组的长度截断或重复。
#4
-1
Reading part of the input data
读取部分输入数据
df <- read.table(text = "name row
a 1
a 2
a 3
a 4
a 5
b 1
b 2
b 3
b 4
b 5
b 6
b 7
b 8
b 9
b 10
b 11
b 12
b 13
b 14
b 15", header = TRUE)
Required output can be achieved using conditional mutate from dplyr package.
使用dplyr包中的条件mutate可以实现所需的输出。
df %>% mutate(newRow = ifelse(row > 11, row - 11, row))
# name row newRow
# 1 a 1 1
# 2 a 2 2
# 3 a 3 3
# 4 a 4 4
# 5 a 5 5
# 6 b 1 1
# 7 b 2 2
# 8 b 3 3
# 9 b 4 4
# 10 b 5 5
# 11 b 6 6
# 12 b 7 7
# 13 b 8 8
# 14 b 9 9
# 15 b 10 10
# 16 b 11 11
# 17 b 12 1
# 18 b 13 2
# 19 b 14 3
# 20 b 15 4
If the row has number more than 22, then the complex statement ((row-1) %% 11 ) + 1 should be used
如果行的数量大于22,则应使用复杂语句((row-1)%% 11)+ 1
df <- data.frame(name=c(rep("a", 5), rep("b", 23)), row=c(1:5,1:23))
df %>% mutate(newRow = ifelse(row > 11, ((row-1) %% 11 ) + 1, row))
# name row newRow
# 1 a 1 1
# 2 a 2 2
# 3 a 3 3
# 4 a 4 4
# 5 a 5 5
# 6 b 1 1
# 7 b 2 2
# 8 b 3 3
# 9 b 4 4
# 10 b 5 5
# 11 b 6 6
# 12 b 7 7
# 13 b 8 8
# 14 b 9 9
# 15 b 10 10
# 16 b 11 11
# 17 b 12 1
# 18 b 13 2
# 19 b 14 3
# 20 b 15 4
# 21 b 16 5
# 22 b 17 6
# 23 b 18 7
# 24 b 19 8
# 25 b 20 9
# 26 b 21 10
# 27 b 22 11
# 28 b 23 1
The same output using data.table,
使用data.table的相同输出,
dt <- data.table(df)
dt[, newRow := ifelse(row > 11, ((row-1) %% 11 ) + 1, row)]
#1
3
Try this, using dplyr:
试试这个,使用dplyr:
df <- data.frame(name=c(rep("a", 5), rep("b", 13)), stringsAsFactors = FALSE)
library(dplyr)
df %>%
group_by(name) %>%
mutate(
row = row_number(),
new_row = (row - 1L) %% 11L + 1L
) %>%
ungroup()
# # A tibble: 18 × 3
# name row new_row
# <chr> <int> <int>
# 1 a 1 1
# 2 a 2 2
# 3 a 3 3
# 4 a 4 4
# 5 a 5 5
# 6 b 1 1
# 7 b 2 2
# 8 b 3 3
# 9 b 4 4
# 10 b 5 5
# 11 b 6 6
# 12 b 7 7
# 13 b 8 8
# 14 b 9 9
# 15 b 10 10
# 16 b 11 11
# 17 b 12 1
# 18 b 13 2
Using base-R:
do.call(
rbind.data.frame,
by(df, df$name,
function(x) within(x, {
row = seq_len(nrow(x))
new_row = (row - 1L) %% 11L + 1L
})
))
# name new_row row
# a.1 a 1 1
# a.2 a 2 2
# a.3 a 3 3
# a.4 a 4 4
# a.5 a 5 5
# b.6 b 1 1
# b.7 b 2 2
# b.8 b 3 3
# b.9 b 4 4
# b.10 b 5 5
# b.11 b 6 6
# b.12 b 7 7
# b.13 b 8 8
# b.14 b 9 9
# b.15 b 10 10
# b.16 b 11 11
# b.17 b 1 12
# b.18 b 2 13
#2
2
Another idea using ave from base R,
使用基础R的ave的另一个想法,
with(df, ave(row, name, FUN = function(i) replace(i, i>11, seq(i[i > 11]))))
#[1] 1 2 3 4 5 1 2 3 4 5 6 7 8 9 10 11 1 2 3 4
Good catch on the limitation and suggestion by @r2evans,
很好地了解了@ r2evans的限制和建议,
with(df, ave(row, name, FUN = function(i) (i-1)%%11+1))
#3
0
Using the base R functions ave and rep, we can do
使用基本R函数ave和rep,我们可以做到
ave(df$row, df$name, FUN=function(x) rep(1:11, length.out=length(x)))
[1] 1 2 3 4 5 1 2 3 4 5 6 7 8 9 10 11 1 2 3 4
This uses with reps length.out argument to repeat the sequence (1:11) truncating or repeating according to the length of each group.
这与reps length.out参数一起使用,重复序列(1:11)根据每个组的长度截断或重复。
#4
-1
Reading part of the input data
读取部分输入数据
df <- read.table(text = "name row
a 1
a 2
a 3
a 4
a 5
b 1
b 2
b 3
b 4
b 5
b 6
b 7
b 8
b 9
b 10
b 11
b 12
b 13
b 14
b 15", header = TRUE)
Required output can be achieved using conditional mutate from dplyr package.
使用dplyr包中的条件mutate可以实现所需的输出。
df %>% mutate(newRow = ifelse(row > 11, row - 11, row))
# name row newRow
# 1 a 1 1
# 2 a 2 2
# 3 a 3 3
# 4 a 4 4
# 5 a 5 5
# 6 b 1 1
# 7 b 2 2
# 8 b 3 3
# 9 b 4 4
# 10 b 5 5
# 11 b 6 6
# 12 b 7 7
# 13 b 8 8
# 14 b 9 9
# 15 b 10 10
# 16 b 11 11
# 17 b 12 1
# 18 b 13 2
# 19 b 14 3
# 20 b 15 4
If the row has number more than 22, then the complex statement ((row-1) %% 11 ) + 1 should be used
如果行的数量大于22,则应使用复杂语句((row-1)%% 11)+ 1
df <- data.frame(name=c(rep("a", 5), rep("b", 23)), row=c(1:5,1:23))
df %>% mutate(newRow = ifelse(row > 11, ((row-1) %% 11 ) + 1, row))
# name row newRow
# 1 a 1 1
# 2 a 2 2
# 3 a 3 3
# 4 a 4 4
# 5 a 5 5
# 6 b 1 1
# 7 b 2 2
# 8 b 3 3
# 9 b 4 4
# 10 b 5 5
# 11 b 6 6
# 12 b 7 7
# 13 b 8 8
# 14 b 9 9
# 15 b 10 10
# 16 b 11 11
# 17 b 12 1
# 18 b 13 2
# 19 b 14 3
# 20 b 15 4
# 21 b 16 5
# 22 b 17 6
# 23 b 18 7
# 24 b 19 8
# 25 b 20 9
# 26 b 21 10
# 27 b 22 11
# 28 b 23 1
The same output using data.table,
使用data.table的相同输出,
dt <- data.table(df)
dt[, newRow := ifelse(row > 11, ((row-1) %% 11 ) + 1, row)]