lag.formula | R Documentation |
Lags a variable using panel id + time identifiers in a formula.
## S3 method for class 'formula'
lag(
x,
k = 1,
data,
time.step = NULL,
fill = NA,
duplicate.method = c("none", "first"),
...
)
lag_fml(
x,
k = 1,
data,
time.step = NULL,
fill = NA,
duplicate.method = c("none", "first"),
...
)
x |
A formula of the type |
k |
An integer giving the number of lags. Default is 1. For leads, just use a negative number. |
data |
Optional, the data.frame in which to evaluate the formula. If not provided, variables will be fetched in the current environment. |
time.step |
The method to compute the lags, default is |
fill |
Scalar. How to fill the observations without defined lead/lag values.
Default is |
duplicate.method |
If several observations have the same id and time values,
then the notion of lag is not defined for them. If |
... |
Not currently used. |
It returns a vector of the same type and length as the variable to be lagged in the formula.
lag_fml()
: Lags a variable using a formula syntax
Laurent Berge
Alternatively, the function panel
changes a data.frame
into a panel from which
the functions l
and f
(creating leads and lags) can be called. Otherwise you can set
the panel 'live' during the estimation using the argument panel.id
(see for example in
the function feols
).
# simple example with an unbalanced panel
base = data.frame(id = rep(1:2, each = 4),
time = c(1, 2, 3, 4, 1, 4, 6, 9), x = 1:8)
base$lag1 = lag(x~id+time, 1, base) # lag 1
base$lead1 = lag(x~id+time, -1, base) # lead 1
base$lag2_fill0 = lag(x~id+time, 2, base, fill = 0)
# with time.step = "consecutive"
base$lag1_consecutive = lag(x~id+time, 1, base, time.step = "consecutive")
# => works for indiv. 2 because 9 (resp. 6) is consecutive to 6 (resp. 4)
base$lag1_within.consecutive = lag(x~id+time, 1, base, time.step = "within")
# => now two consecutive years within each indiv is one lag
print(base)
# Argument time.step = "consecutive" is
# mostly useful when the time variable is not a number:
# e.g. c("1991q1", "1991q2", "1991q3") etc
# with duplicates
base_dup = data.frame(id = rep(1:2, each = 4),
time = c(1, 1, 1, 2, 1, 2, 2, 3), x = 1:8)
# Error because of duplicate values for (id, time)
try(lag(x~id+time, 1, base_dup))
# Error is bypassed, lag corresponds to first occurence of (id, time)
lag(x~id+time, 1, base_dup, duplicate.method = "first")
# Playing with time steps
base = data.frame(id = rep(1:2, each = 4),
time = c(1, 2, 3, 4, 1, 4, 6, 9), x = 1:8)
# time step: 0.5 (here equivalent to lag of 1)
lag(x~id+time, 2, base, time.step = 0.5)
# Error: wrong time step
try(lag(x~id+time, 2, base, time.step = 7))
# Adding NAs + unsorted IDs
base = data.frame(id = rep(1:2, each = 4),
time = c(4, NA, 3, 1, 2, NA, 1, 3), x = 1:8)
base$lag1 = lag(x~id+time, 1, base)
base$lag1_within = lag(x~id+time, 1, base, time.step = "w")
base_bis = base[order(base$id, base$time),]
print(base_bis)
# You can create variables without specifying the data within data.table:
if(require("data.table")){
base = data.table(id = rep(1:2, each = 3), year = 1990 + rep(1:3, 2), x = 1:6)
base[, x.l1 := lag(x~id+year, 1)]
}
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