R - Mean values of numbers in a sequence in a dataframe column

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I am trying to calculate the mean of values in a column (y) that are being sequentially counted (1, 2, 3, etc.) within another column (x). An example dataframe is shown below.

> df
   x  y
1  1 15
2  2 20
3  4 16
4  5 12
5  6 17
6  8 14
7  9 13
8 10 19

I would like to get a vector result that will read out the mean values of numbers in a numerical sequence. The desired vector would read: 17.5 15 15.33333

I am not sure the best way to produce this desired result, but I have tried and failed to make a for loop using diff(df[,1]) to find the breakpoints.

Any help anyone could provide would be appreciated. This is a small example dataset, but the goal is to apply it to a large dataset.

mean dataframe sequence r
2 Answers

Create a grouping column from the diff using cumsum on a logical vector

with(df, tapply(y, cumsum(c(TRUE, diff(x) != 1)), FUN = mean))

-output

  1        2        3 
17.50000 15.00000 15.33333

data

df <- structure(list(x = c(1L, 2L, 4L, 5L, 6L, 8L, 9L, 10L), y = c(15L, 
20L, 16L, 12L, 17L, 14L, 13L, 19L)), class = "data.frame",
 row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8"))
user_37322710

Update: code reduced 2 lines:

df %>% 
  group_by(id_Group =cumsum(x-lag(x, default = x[1])>=2)) %>% 
  summarise(mean = mean(y, na.rm=TRUE)) %>% 
  pull(mean)

Here is a dplyr version:

  1. calculate the difference between the lagged x
  2. create group with cumsum(diff>=2)
  3. calculate mean and pull the vector.
library(dplyr)

df %>% 
  mutate(diff= x-lag(x, default = x[1])) %>% 
  group_by(id_Group =cumsum(diff>=2)) %>% 
  mutate(mean = mean(y, na.rm=TRUE)) %>% 
  slice(1) %>% 
  pull(mean)

[1] 17.50000 15.00000 15.33333
user_133216470
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