将列添加到数据框列表并进行增量添加/循环通过 df 进行简单添加

Question

I have this data frame (actually, a list of those dfs):

ALL <- data.frame(x = 1:3, y = c("a", "b", "c"))

I want to add a column which adds up a certain value until the end of the data frame, like this: 0 + 0.05 = 0.05, 0.05 + 0.05 = 0.1, 0.1 + 0.05 = 0.15 and so on.

So, in my example the result would be

ALL <- data.frame(x = 1:3, y = c("a", "b", "c"), z=(0,0.05,0.1)

I guess the way to go would be using cbind with lapply (assuming ALL is a list of dfs): ALL<- lapply(ALL, function(x) cbind(x, z = ???))

But my brain simple doesn't come up with the right formula for z.

Your help is very appreciated.

Greetings

trumfnator

Answer 1

We can use transform to add new column with seq to generate the sequence.

lapply(list_df, function(df) transform(df, z = seq(0, by = 0.05, length.out = nrow(df))))

#[[1]]
#  x y    z
#1 1 a 0.00
#2 2 b 0.05
#3 3 c 0.10

#[[2]]
#  x y    z
#1 1 a 0.00
#2 2 b 0.05
#3 3 c 0.10
#4 4 d 0.15

---

In tidyverse we can do the same by

library(dplyr)
library(purrr)

map(list_df, ~.x %>% mutate(z = seq(0, by = 0.05, length.out = n())))

data

ALL <- data.frame(x = 1:3, y = c("a", "b", "c"))
ALL1 <- data.frame(x = 1:4, y = c("a", "b", "c", "d"))
list_df <- list(ALL, ALL1)

Answer 2

也许这就是你想要的

ALL$z <- 0.05*(0:(nrow(ALL)-1))

Answer 3

We can do an assignment and then return the dataset

lapply(list_df, function(x) {x$z <- seq(0, by= 0.05, length.out = nrow(x));x})

data

ALL <- data.frame(x = 1:3, y = c("a", "b", "c"))
ALL1 <- data.frame(x = 1:4, y = c("a", "b", "c", "d"))
list_df <- list(ALL, ALL1)