Return full array if one row within the array contains a given value R [duplicate]

Question

I have nested data that looks like this:

ID  Date Behavior
1   1    FALSE
1   2    TRUE
1   3    TRUE
2   1    TRUE
2   2    FALSE
3   1    TRUE
3   2    TRUE

I'd like to return each array of values for a given ID that contains at least one occurrence of FALSE. I am expecting ID 1 and ID 2 to be returned, with each row of present data (3 rows for ID 1 and 2 rows for ID2).

EDIT: here is what I am expecting:

ID  Date Behavior
1   1    FALSE
1   2    TRUE
1   3    TRUE
2   1    TRUE
2   2    FALSE

I'm wondering if this is a for loop or a while function - any and all help is appreciated...

Extra points for python code that mimics the R code!

Answer 1

Here's a possible data.table approach (assuming df is your data set)

library(data.table)
setDT(df)[, .SD[any(!Behavior)], ID] # you can also replace any(!Behavior) with !all(Behavior)
#    ID Date Behavior
# 1:  1    1    FALSE
# 2:  1    2     TRUE
# 3:  1    3     TRUE
# 4:  2    1     TRUE
# 5:  2    2    FALSE

Edit: a bit more efficient solution by @Arun

setDT(df)[, if (any(!Behavior)) .SD, ID]

---

Or a similar dplyr approach

library(dplyr)
df %>%
  group_by(ID) %>%
  filter(any(!Behavior))

# Source: local data table [5 x 3]
# Groups: ID
# 
#   ID Date Behavior
# 1  1    1    FALSE
# 2  1    2     TRUE
# 3  1    3     TRUE
# 4  2    1     TRUE
# 5  2    2    FALSE