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generate bootstrap sample dependent on column
How to sort a dataframe by multiple column(s)?Resample a longitudinal dataset in Rdata.table vs dplyr: can one do something well the other can't or does poorly?Using dplyr's do to perform bootstrap replicationsHow to resample data by clusters (block sampling) with replacement in R using Sampling packageReproducable bootstrap samplesreplicate the row depending on a columnBootstrapping on R - resampling for a stratfied sampleResampling cross-sectional time series data in RHow to subsample different numbers by ID and bootstrap in R
I have a data set like this
set.seed(1)
df <- data.frame(ID = rep(1:4, each = 3),
x = c(1,2,3,2,3,4,1,2,3,3,4,5),
V1 = rnorm(12))
> df
ID x V1
1 1 1 -0.6264538
2 1 2 0.1836433
3 1 3 -0.8356286
4 2 2 1.5952808
5 2 3 0.3295078
6 2 4 -0.8204684
7 3 1 0.4874291
8 3 2 0.7383247
9 3 3 0.5757814
10 4 3 -0.3053884
11 4 4 1.5117812
12 4 5 0.3898432
this example contains 4 individuals, defined by ID.
Each individual has an observation period x. For example ID 1 is observed at time points 1,2,3.
In this example I have 2 observations at time point 1 (ID 1 and ID 3),
and 3 observations at time point 2 (IDs 1,2,3)
I now want a bootstrapped (sample with replacement) data set that contains the same number of observations at each time point.
In this example the data set could look like this:
> df
ID x V1
1 1 1 -0.6264538
1 1 1 -0.6264538
2 1 2 0.1836433
2 1 2 0.1836433
3 1 3 -0.8356286
4 2 2 1.5952808
5 2 3 0.3295078
6 2 4 -0.8204684
6 2 4 -0.8204684
7 3 1 0.4874291
7 3 1 0.4874291
8 3 2 0.7383247
9 3 3 0.5757814
10 4 3 -0.3053884
11 4 4 1.5117812
11 4 4 1.5117812
12 4 5 0.3898432
12 4 5 0.3898432
12 4 5 0.3898432
12 4 5 0.3898432
this data set now has 4 observations at each time point.
r dplyr sampling resampling
asked Mar 8 at 7:50
spore234spore234
1,34712050
add a comment |
I have a data set like this
set.seed(1)
df <- data.frame(ID = rep(1:4, each = 3),
x = c(1,2,3,2,3,4,1,2,3,3,4,5),
V1 = rnorm(12))
> df
ID x V1
1 1 1 -0.6264538
2 1 2 0.1836433
3 1 3 -0.8356286
4 2 2 1.5952808
5 2 3 0.3295078
6 2 4 -0.8204684
7 3 1 0.4874291
8 3 2 0.7383247
9 3 3 0.5757814
10 4 3 -0.3053884
11 4 4 1.5117812
12 4 5 0.3898432
this example contains 4 individuals, defined by ID.
Each individual has an observation period x. For example ID 1 is observed at time points 1,2,3.
In this example I have 2 observations at time point 1 (ID 1 and ID 3),
and 3 observations at time point 2 (IDs 1,2,3)
I now want a bootstrapped (sample with replacement) data set that contains the same number of observations at each time point.
In this example the data set could look like this:
> df
ID x V1
1 1 1 -0.6264538
1 1 1 -0.6264538
2 1 2 0.1836433
2 1 2 0.1836433
3 1 3 -0.8356286
4 2 2 1.5952808
5 2 3 0.3295078
6 2 4 -0.8204684
6 2 4 -0.8204684
7 3 1 0.4874291
7 3 1 0.4874291
8 3 2 0.7383247
9 3 3 0.5757814
10 4 3 -0.3053884
11 4 4 1.5117812
11 4 4 1.5117812
12 4 5 0.3898432
12 4 5 0.3898432
12 4 5 0.3898432
12 4 5 0.3898432
this data set now has 4 observations at each time point.
r dplyr sampling resampling
asked Mar 8 at 7:50
spore234spore234
1,34712050
add a comment |
I have a data set like this
set.seed(1)
df <- data.frame(ID = rep(1:4, each = 3),
x = c(1,2,3,2,3,4,1,2,3,3,4,5),
V1 = rnorm(12))
> df
ID x V1
1 1 1 -0.6264538
2 1 2 0.1836433
3 1 3 -0.8356286
4 2 2 1.5952808
5 2 3 0.3295078
6 2 4 -0.8204684
7 3 1 0.4874291
8 3 2 0.7383247
9 3 3 0.5757814
10 4 3 -0.3053884
11 4 4 1.5117812
12 4 5 0.3898432
this example contains 4 individuals, defined by ID.
Each individual has an observation period x. For example ID 1 is observed at time points 1,2,3.
In this example I have 2 observations at time point 1 (ID 1 and ID 3),
and 3 observations at time point 2 (IDs 1,2,3)
I now want a bootstrapped (sample with replacement) data set that contains the same number of observations at each time point.
In this example the data set could look like this:
> df
ID x V1
1 1 1 -0.6264538
1 1 1 -0.6264538
2 1 2 0.1836433
2 1 2 0.1836433
3 1 3 -0.8356286
4 2 2 1.5952808
5 2 3 0.3295078
6 2 4 -0.8204684
6 2 4 -0.8204684
7 3 1 0.4874291
7 3 1 0.4874291
8 3 2 0.7383247
9 3 3 0.5757814
10 4 3 -0.3053884
11 4 4 1.5117812
11 4 4 1.5117812
12 4 5 0.3898432
12 4 5 0.3898432
12 4 5 0.3898432
12 4 5 0.3898432
this data set now has 4 observations at each time point.
r dplyr sampling resampling
asked Mar 8 at 7:50
spore234spore234
1,34712050
I have a data set like this
set.seed(1)
df <- data.frame(ID = rep(1:4, each = 3),
x = c(1,2,3,2,3,4,1,2,3,3,4,5),
V1 = rnorm(12))
> df
ID x V1
1 1 1 -0.6264538
2 1 2 0.1836433
3 1 3 -0.8356286
4 2 2 1.5952808
5 2 3 0.3295078
6 2 4 -0.8204684
7 3 1 0.4874291
8 3 2 0.7383247
9 3 3 0.5757814
10 4 3 -0.3053884
11 4 4 1.5117812
12 4 5 0.3898432
this example contains 4 individuals, defined by ID.
Each individual has an observation period x. For example ID 1 is observed at time points 1,2,3.
In this example I have 2 observations at time point 1 (ID 1 and ID 3),
and 3 observations at time point 2 (IDs 1,2,3)
I now want a bootstrapped (sample with replacement) data set that contains the same number of observations at each time point.
In this example the data set could look like this:
> df
ID x V1
1 1 1 -0.6264538
1 1 1 -0.6264538
2 1 2 0.1836433
2 1 2 0.1836433
3 1 3 -0.8356286
4 2 2 1.5952808
5 2 3 0.3295078
6 2 4 -0.8204684
6 2 4 -0.8204684
7 3 1 0.4874291
7 3 1 0.4874291
8 3 2 0.7383247
9 3 3 0.5757814
10 4 3 -0.3053884
11 4 4 1.5117812
11 4 4 1.5117812
12 4 5 0.3898432
12 4 5 0.3898432
12 4 5 0.3898432
12 4 5 0.3898432
this data set now has 4 observations at each time point.
r dplyr sampling resampling
r dplyr sampling resampling
asked Mar 8 at 7:50
spore234spore234
1,34712050
asked Mar 8 at 7:50
spore234spore234
1,34712050
asked Mar 8 at 7:50
spore234spore234
1,34712050
asked Mar 8 at 7:50
spore234spore234
1,34712050
asked Mar 8 at 7:50
spore234spore234
1,34712050
1,34712050
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
We could first find the maximum number of times x occurs and sample_n for each x with replace = TRUE to get equal number of rows for each x.
max_sample <- max(table(df$x))
library(dplyr)
df %>%
group_by(x) %>%
sample_n(max_sample, replace = TRUE) %>%
arrange(x)
# ID x V1
# <int> <dbl> <dbl>
# 1 3 1 0.487
# 2 1 1 -0.626
# 3 1 1 -0.626
# 4 1 1 -0.626
# 5 3 2 0.738
# 6 2 2 1.60
# 7 2 2 1.60
# 8 3 2 0.738
# 9 4 3 -0.305
#10 2 3 0.330
#11 2 3 0.330
#12 4 3 -0.305
#13 4 4 1.51
#14 4 4 1.51
#15 4 4 1.51
#16 4 4 1.51
#17 4 5 0.390
#18 4 5 0.390
#19 4 5 0.390
#20 4 5 0.390
answered Mar 8 at 8:02
Ronak ShahRonak Shah
42.8k104266
thank, I should add thatxdoes not always start with 1, it ranges from -20 to +20
– spore234
Mar 8 at 8:06
@spore234 umm...it should not matter I think because we are counting frequency ofxwithtableirrespective of what it's value is.
– Ronak Shah
Mar 8 at 8:08
add a comment |
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1 Answer
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1 Answer
1
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oldest
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oldest
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active
oldest
votes
We could first find the maximum number of times x occurs and sample_n for each x with replace = TRUE to get equal number of rows for each x.
max_sample <- max(table(df$x))
library(dplyr)
df %>%
group_by(x) %>%
sample_n(max_sample, replace = TRUE) %>%
arrange(x)
# ID x V1
# <int> <dbl> <dbl>
# 1 3 1 0.487
# 2 1 1 -0.626
# 3 1 1 -0.626
# 4 1 1 -0.626
# 5 3 2 0.738
# 6 2 2 1.60
# 7 2 2 1.60
# 8 3 2 0.738
# 9 4 3 -0.305
#10 2 3 0.330
#11 2 3 0.330
#12 4 3 -0.305
#13 4 4 1.51
#14 4 4 1.51
#15 4 4 1.51
#16 4 4 1.51
#17 4 5 0.390
#18 4 5 0.390
#19 4 5 0.390
#20 4 5 0.390
answered Mar 8 at 8:02
Ronak ShahRonak Shah
42.8k104266
thank, I should add thatxdoes not always start with 1, it ranges from -20 to +20
– spore234
Mar 8 at 8:06
@spore234 umm...it should not matter I think because we are counting frequency ofxwithtableirrespective of what it's value is.
– Ronak Shah
Mar 8 at 8:08
add a comment |
We could first find the maximum number of times x occurs and sample_n for each x with replace = TRUE to get equal number of rows for each x.
max_sample <- max(table(df$x))
library(dplyr)
df %>%
group_by(x) %>%
sample_n(max_sample, replace = TRUE) %>%
arrange(x)
# ID x V1
# <int> <dbl> <dbl>
# 1 3 1 0.487
# 2 1 1 -0.626
# 3 1 1 -0.626
# 4 1 1 -0.626
# 5 3 2 0.738
# 6 2 2 1.60
# 7 2 2 1.60
# 8 3 2 0.738
# 9 4 3 -0.305
#10 2 3 0.330
#11 2 3 0.330
#12 4 3 -0.305
#13 4 4 1.51
#14 4 4 1.51
#15 4 4 1.51
#16 4 4 1.51
#17 4 5 0.390
#18 4 5 0.390
#19 4 5 0.390
#20 4 5 0.390
answered Mar 8 at 8:02
Ronak ShahRonak Shah
42.8k104266
thank, I should add thatxdoes not always start with 1, it ranges from -20 to +20
– spore234
Mar 8 at 8:06
@spore234 umm...it should not matter I think because we are counting frequency ofxwithtableirrespective of what it's value is.
– Ronak Shah
Mar 8 at 8:08
add a comment |
We could first find the maximum number of times x occurs and sample_n for each x with replace = TRUE to get equal number of rows for each x.
max_sample <- max(table(df$x))
library(dplyr)
df %>%
group_by(x) %>%
sample_n(max_sample, replace = TRUE) %>%
arrange(x)
# ID x V1
# <int> <dbl> <dbl>
# 1 3 1 0.487
# 2 1 1 -0.626
# 3 1 1 -0.626
# 4 1 1 -0.626
# 5 3 2 0.738
# 6 2 2 1.60
# 7 2 2 1.60
# 8 3 2 0.738
# 9 4 3 -0.305
#10 2 3 0.330
#11 2 3 0.330
#12 4 3 -0.305
#13 4 4 1.51
#14 4 4 1.51
#15 4 4 1.51
#16 4 4 1.51
#17 4 5 0.390
#18 4 5 0.390
#19 4 5 0.390
#20 4 5 0.390
answered Mar 8 at 8:02
Ronak ShahRonak Shah
42.8k104266
We could first find the maximum number of times x occurs and sample_n for each x with replace = TRUE to get equal number of rows for each x.
max_sample <- max(table(df$x))
library(dplyr)
df %>%
group_by(x) %>%
sample_n(max_sample, replace = TRUE) %>%
arrange(x)
# ID x V1
# <int> <dbl> <dbl>
# 1 3 1 0.487
# 2 1 1 -0.626
# 3 1 1 -0.626
# 4 1 1 -0.626
# 5 3 2 0.738
# 6 2 2 1.60
# 7 2 2 1.60
# 8 3 2 0.738
# 9 4 3 -0.305
#10 2 3 0.330
#11 2 3 0.330
#12 4 3 -0.305
#13 4 4 1.51
#14 4 4 1.51
#15 4 4 1.51
#16 4 4 1.51
#17 4 5 0.390
#18 4 5 0.390
#19 4 5 0.390
#20 4 5 0.390
answered Mar 8 at 8:02
Ronak ShahRonak Shah
42.8k104266
answered Mar 8 at 8:02
Ronak ShahRonak Shah
42.8k104266
answered Mar 8 at 8:02
Ronak ShahRonak Shah
42.8k104266
answered Mar 8 at 8:02
Ronak ShahRonak Shah
42.8k104266
42.8k104266
thank, I should add thatxdoes not always start with 1, it ranges from -20 to +20
– spore234
Mar 8 at 8:06
@spore234 umm...it should not matter I think because we are counting frequency ofxwithtableirrespective of what it's value is.
– Ronak Shah
Mar 8 at 8:08
add a comment |
thank, I should add thatxdoes not always start with 1, it ranges from -20 to +20
– spore234
Mar 8 at 8:06
@spore234 umm...it should not matter I think because we are counting frequency ofxwithtableirrespective of what it's value is.
– Ronak Shah
Mar 8 at 8:08
thank, I should add that
x does not always start with 1, it ranges from -20 to +20– spore234
Mar 8 at 8:06
thank, I should add that
x does not always start with 1, it ranges from -20 to +20– spore234
Mar 8 at 8:06
@spore234 umm...it should not matter I think because we are counting frequency of
x with table irrespective of what it's value is.– Ronak Shah
Mar 8 at 8:08
@spore234 umm...it should not matter I think because we are counting frequency of
x with table irrespective of what it's value is.– Ronak Shah
Mar 8 at 8:08
add a comment |
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