Given this tibble:
tibble(x = c(1:9))
I want to add a column x_lag_1 = c(NA,1:8), a column x_lag_2 = c(NA,NA,1:7), etc.
Up to x_lag_n.
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GiulioGCantone
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3 Answers
3
This can be quick with data.table:
library(data.table)
n <- seq(4)
setDT(df)[, paste0('x_lag_', n) := shift(x, n)]
df
x x_lag_1 x_lag_2 x_lag_3 x_lag_4
1: 1 NA NA NA NA
2: 2 1 NA NA NA
3: 3 2 1 NA NA
4: 4 3 2 1 NA
5: 5 4 3 2 1
6: 6 5 4 3 2
7: 7 6 5 4 3
8: 8 7 6 5 4
9: 9 8 7 6 5
Onyambu
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You may use map_dfc to add n new columns.
library(dplyr)
library(purrr)
df <- tibble(x = c(1:9))
n <- 3
bind_cols(df, map_dfc(seq_len(n), ~df %>%
transmute(!!paste0('x_lag', .x) := lag(x, .x))))
# x x_lag1 x_lag2 x_lag3
# <int> <int> <int> <int>
#1 1 NA NA NA
#2 2 1 NA NA
#3 3 2 1 NA
#4 4 3 2 1
#5 5 4 3 2
#6 6 5 4 3
#7 7 6 5 4
#8 8 7 6 5
#9 9 8 7 6
Ronak Shah
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is there a reason why this formula does not respect groups in the df, or a way to enforce them? – GiulioGCantone Dec 29 '21 at 23:53
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No, it does not work properly, you can check it. ` df <- tibble(a = c(1:9), y = c(rep("A",5),rep("B",4)) ) %>% group_by(y)` `bind_cols(df, map_dfc(seq(2), ~df %>% transmute(!!paste0('a_lag', .x) := dplyr::lag(a, .x))))` – GiulioGCantone Dec 30 '21 at 01:35
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The result is exactly the same, you can check it. – GiulioGCantone Dec 30 '21 at 01:42
0
Edit 2: Reworked the answer to contemplate the case of a grouped df.
library(tidyverse)
set.seed(123)
df <- tibble(group = sample(letters[1:3], 30, replace = TRUE), x = c(1:30))
formulas <- seq(3, 12, 3) %>%
map(~ as.formula(str_glue("~lag(.,n={.x})"))) %>%
set_names(str_c("lag", seq(3, 12, 3)))
df %>%
summarise(x, across(x, lst(!!!formulas)))
#> # A tibble: 30 × 5
#> x x_lag3 x_lag6 x_lag9 x_lag12
#> <int> <int> <int> <int> <int>
#> 1 1 NA NA NA NA
#> 2 2 NA NA NA NA
#> 3 3 NA NA NA NA
#> 4 4 1 NA NA NA
#> 5 5 2 NA NA NA
#> 6 6 3 NA NA NA
#> 7 7 4 1 NA NA
#> 8 8 5 2 NA NA
#> 9 9 6 3 NA NA
#> 10 10 7 4 1 NA
#> # … with 20 more rows
df %>%
group_by(group) %>%
summarise(x, across(x, lst(!!!formulas)), .groups = "drop")
#> # A tibble: 30 × 6
#> group x x_lag3 x_lag6 x_lag9 x_lag12
#> <chr> <int> <int> <int> <int> <int>
#> 1 a 10 NA NA NA NA
#> 2 a 13 NA NA NA NA
#> 3 a 16 NA NA NA NA
#> 4 a 19 10 NA NA NA
#> 5 a 20 13 NA NA NA
#> 6 a 21 16 NA NA NA
#> 7 a 22 19 10 NA NA
#> 8 a 27 20 13 NA NA
#> 9 b 4 NA NA NA NA
#> 10 b 6 NA NA NA NA
#> # … with 20 more rows
Created on 2021-12-30 by the reprex package (v2.0.1)
jpdugo17
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This works pretty bad when instead of `1:n` there is a sequence of integers through a `by=`. – GiulioGCantone Dec 30 '21 at 00:02
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```df <- tibble(x = c(1:9)) walk(str_c("x_lag_", seq(6,21,by=3)), ~ { df[[.]] <<- dplyr::lag(df$x, parse_number(.)) }) ``` – GiulioGCantone Dec 30 '21 at 00:32
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You are lagging by 6 positions , then 9, but the dataset contains only 9 rows. Do you want to preserve the names but lag by a different number? – jpdugo17 Dec 30 '21 at 00:46
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Is there a way to make this code respect groups? `group_walk` gives this error: `Error in UseMethod("group_map") : no applicable method for 'group_map' applied to an object of class "character"` – GiulioGCantone Dec 30 '21 at 01:04
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Is `df` supposed to be grouped by a variable? Where did the groups came from? – jpdugo17 Dec 30 '21 at 01:10
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@GiulioGCantone I edited the answer to include a case when the df is grouped and what it would look like. – jpdugo17 Dec 30 '21 at 01:42
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It worked, with 2 workaround: - addition in `df` `bind_cols(df,group, map_dfc(seq_len(n))`, this keeps the existence of other columns - `relocate` group, which is just to make it tidier. However, I will wait a bunch of days because I feel that this is the "wrong way" to basically make a "mass mutate". – GiulioGCantone Dec 30 '21 at 02:05