dplyr::arrange refuse to arrange dataframe by timestamp column?
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I have the following dataframe (here is the sample):
df <- structure(list(user_id = c(1L, 1L, 1L, 1L, 1L, 1L), obs_id = c(1L,
2L, 2L, 2L, 2L, 2L), scroll_id = c(3L, 1L, 2L, 3L, 4L, 5L), timestamp = c(-1.74966971796047,
-1.70403832189443, -1.70379906928687, -1.70361867040459, -1.70347088963619,
-1.70319128699835), row_num = 1:6, scroll_length = c(6, 9, 14,
12, 13, 26), x_mean = c(-1.74134749014902, -1.19087086808828,
1.36178725012622, -1.32786301490502, 1.24184201608646, -1.31953110973881
), y_mean = c(-4.93507461932646, 0.0304680987883223, 0.140001980341645,
0.61911843405746, 0.434230282460559, 0.438563278736709), dx_mean = c(-0.514034686928457,
-0.709482080612108, 0.924636289935977, -0.702980646737082, 0.515080876392673,
-0.359676884238743), dy_mean = c(0.972265996197407, -0.692113718739584,
-0.162463490249733, -0.373682612876388, -0.0663766957581004,
0.293619375985922)), .Names = c("user_id", "obs_id", "scroll_id",
"timestamp", "row_num", "scroll_length", "x_mean", "y_mean",
"dx_mean", "dy_mean"), row.names = c(NA, -6L), class = c("tbl_df",
"tbl", "data.frame"))
I want to arrange by timestamp column but I get the following error:
data %>% arrange(timestamp)
data %>% arrange("timestamp")
Error in arrange_impl(.data, dots) : Argument 1 is of unsupported type
matrix
Please advise how to make it work. I know that timestamp is a function and matrix but here it's a column and I "want" dplyr
to "understand" that it's a column.
As @sotos asked:
sessionInfo():
R version 3.4.4 (2018-03-15)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 18.04.1 LTS
Matrix products: default
BLAS: /usr/lib/x86_64-linux-gnu/blas/libblas.so.3.7.1
LAPACK: /usr/lib/x86_64-linux-gnu/lapack/liblapack.so.3.7.1
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C LC_TIME=en_IL.UTF-8 LC_COLLATE=en_US.UTF-8 LC_MONETARY=en_IL.UTF-8
[6] LC_MESSAGES=en_US.UTF-8 LC_PAPER=en_IL.UTF-8 LC_NAME=C LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_IL.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] parallel stats graphics grDevices utils datasets methods base
other attached packages:
[1] bindrcpp_0.2.2 rebus_0.1-3 philentropy_0.2.0 reshape2_1.4.3 broom_0.5.0 dummies_1.5.6 hms_0.4.2
[8] anytime_0.3.1 data.table_1.11.8 bit64_0.9-7 bit_1.1-14 car_3.0-2 carData_3.0-2 caret_6.0-80
[15] lattice_0.20-35 xgboost_0.71.2 doSNOW_1.0.16 snow_0.4-3 doMC_1.3.5 iterators_1.0.10 foreach_1.4.4
[22] randomForest_4.6-14 htmlwidgets_1.3 plotly_4.8.0 jsonlite_1.5 pROC_1.13.0 knitr_1.20 lubridate_1.7.4
[29] MASS_7.3-49 chron_2.3-53 forcats_0.3.0 stringr_1.3.1 dplyr_0.7.7 purrr_0.2.5 readr_1.1.1
[36] tidyr_0.8.2 tibble_1.4.2 ggplot2_3.1.0 tidyverse_1.2.1
loaded via a namespace (and not attached):
[1] nlme_3.1-137 dimRed_0.1.0 httr_1.3.1 tools_3.4.4 backports_1.1.2 R6_2.3.0
[7] rpart_4.1-13 rebus.base_0.0-3 lazyeval_0.2.1 colorspace_1.3-2 nnet_7.3-12 withr_2.1.2
[13] tidyselect_0.2.5 curl_3.1 compiler_3.4.4 cli_1.0.1 rvest_0.3.2 xml2_1.2.0
[19] scales_1.0.0 sfsmisc_1.1-2 DEoptimR_1.0-8 robustbase_0.93-3 RApiDatetime_0.0.4 digest_0.6.18
[25] rebus.unicode_0.0-2 foreign_0.8-70 rio_0.5.10 pkgconfig_2.0.2 htmltools_0.3.6 rlang_0.3.0.1
[31] readxl_1.1.0 ddalpha_1.3.4 rstudioapi_0.8 bindr_0.1.1 zip_1.0.0 ModelMetrics_1.2.0
[37] magrittr_1.5 Matrix_1.2-14 Rcpp_0.12.19 munsell_0.5.0 abind_1.4-5 stringi_1.2.4
[43] yaml_2.2.0 plyr_1.8.4 recipes_0.1.3 grid_3.4.4 pls_2.7-0 crayon_1.3.4
[49] rebus.datetimes_0.0-1 haven_1.1.2 splines_3.4.4 pillar_1.3.0 rebus.numbers_0.0-1 codetools_0.2-15
[55] stats4_3.4.4 CVST_0.2-2 magic_1.5-9 glue_1.3.0 modelr_0.1.2 cellranger_1.1.0
[61] gtable_0.2.0 kernlab_0.9-27 assertthat_0.2.0 DRR_0.0.3 openxlsx_4.1.0 gower_0.1.2
[67] prodlim_2018.04.18 class_7.3-14 survival_2.42-3 viridisLite_0.3.0 geometry_0.3-6 timeDate_3043.102
[73] RcppRoll_0.3.0 lava_1.6.3 ipred_0.9-7
r dataframe dplyr tidyverse
add a comment |
up vote
1
down vote
favorite
I have the following dataframe (here is the sample):
df <- structure(list(user_id = c(1L, 1L, 1L, 1L, 1L, 1L), obs_id = c(1L,
2L, 2L, 2L, 2L, 2L), scroll_id = c(3L, 1L, 2L, 3L, 4L, 5L), timestamp = c(-1.74966971796047,
-1.70403832189443, -1.70379906928687, -1.70361867040459, -1.70347088963619,
-1.70319128699835), row_num = 1:6, scroll_length = c(6, 9, 14,
12, 13, 26), x_mean = c(-1.74134749014902, -1.19087086808828,
1.36178725012622, -1.32786301490502, 1.24184201608646, -1.31953110973881
), y_mean = c(-4.93507461932646, 0.0304680987883223, 0.140001980341645,
0.61911843405746, 0.434230282460559, 0.438563278736709), dx_mean = c(-0.514034686928457,
-0.709482080612108, 0.924636289935977, -0.702980646737082, 0.515080876392673,
-0.359676884238743), dy_mean = c(0.972265996197407, -0.692113718739584,
-0.162463490249733, -0.373682612876388, -0.0663766957581004,
0.293619375985922)), .Names = c("user_id", "obs_id", "scroll_id",
"timestamp", "row_num", "scroll_length", "x_mean", "y_mean",
"dx_mean", "dy_mean"), row.names = c(NA, -6L), class = c("tbl_df",
"tbl", "data.frame"))
I want to arrange by timestamp column but I get the following error:
data %>% arrange(timestamp)
data %>% arrange("timestamp")
Error in arrange_impl(.data, dots) : Argument 1 is of unsupported type
matrix
Please advise how to make it work. I know that timestamp is a function and matrix but here it's a column and I "want" dplyr
to "understand" that it's a column.
As @sotos asked:
sessionInfo():
R version 3.4.4 (2018-03-15)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 18.04.1 LTS
Matrix products: default
BLAS: /usr/lib/x86_64-linux-gnu/blas/libblas.so.3.7.1
LAPACK: /usr/lib/x86_64-linux-gnu/lapack/liblapack.so.3.7.1
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C LC_TIME=en_IL.UTF-8 LC_COLLATE=en_US.UTF-8 LC_MONETARY=en_IL.UTF-8
[6] LC_MESSAGES=en_US.UTF-8 LC_PAPER=en_IL.UTF-8 LC_NAME=C LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_IL.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] parallel stats graphics grDevices utils datasets methods base
other attached packages:
[1] bindrcpp_0.2.2 rebus_0.1-3 philentropy_0.2.0 reshape2_1.4.3 broom_0.5.0 dummies_1.5.6 hms_0.4.2
[8] anytime_0.3.1 data.table_1.11.8 bit64_0.9-7 bit_1.1-14 car_3.0-2 carData_3.0-2 caret_6.0-80
[15] lattice_0.20-35 xgboost_0.71.2 doSNOW_1.0.16 snow_0.4-3 doMC_1.3.5 iterators_1.0.10 foreach_1.4.4
[22] randomForest_4.6-14 htmlwidgets_1.3 plotly_4.8.0 jsonlite_1.5 pROC_1.13.0 knitr_1.20 lubridate_1.7.4
[29] MASS_7.3-49 chron_2.3-53 forcats_0.3.0 stringr_1.3.1 dplyr_0.7.7 purrr_0.2.5 readr_1.1.1
[36] tidyr_0.8.2 tibble_1.4.2 ggplot2_3.1.0 tidyverse_1.2.1
loaded via a namespace (and not attached):
[1] nlme_3.1-137 dimRed_0.1.0 httr_1.3.1 tools_3.4.4 backports_1.1.2 R6_2.3.0
[7] rpart_4.1-13 rebus.base_0.0-3 lazyeval_0.2.1 colorspace_1.3-2 nnet_7.3-12 withr_2.1.2
[13] tidyselect_0.2.5 curl_3.1 compiler_3.4.4 cli_1.0.1 rvest_0.3.2 xml2_1.2.0
[19] scales_1.0.0 sfsmisc_1.1-2 DEoptimR_1.0-8 robustbase_0.93-3 RApiDatetime_0.0.4 digest_0.6.18
[25] rebus.unicode_0.0-2 foreign_0.8-70 rio_0.5.10 pkgconfig_2.0.2 htmltools_0.3.6 rlang_0.3.0.1
[31] readxl_1.1.0 ddalpha_1.3.4 rstudioapi_0.8 bindr_0.1.1 zip_1.0.0 ModelMetrics_1.2.0
[37] magrittr_1.5 Matrix_1.2-14 Rcpp_0.12.19 munsell_0.5.0 abind_1.4-5 stringi_1.2.4
[43] yaml_2.2.0 plyr_1.8.4 recipes_0.1.3 grid_3.4.4 pls_2.7-0 crayon_1.3.4
[49] rebus.datetimes_0.0-1 haven_1.1.2 splines_3.4.4 pillar_1.3.0 rebus.numbers_0.0-1 codetools_0.2-15
[55] stats4_3.4.4 CVST_0.2-2 magic_1.5-9 glue_1.3.0 modelr_0.1.2 cellranger_1.1.0
[61] gtable_0.2.0 kernlab_0.9-27 assertthat_0.2.0 DRR_0.0.3 openxlsx_4.1.0 gower_0.1.2
[67] prodlim_2018.04.18 class_7.3-14 survival_2.42-3 viridisLite_0.3.0 geometry_0.3-6 timeDate_3043.102
[73] RcppRoll_0.3.0 lava_1.6.3 ipred_0.9-7
r dataframe dplyr tidyverse
Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Nov 8 at 11:21
add a comment |
up vote
1
down vote
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up vote
1
down vote
favorite
I have the following dataframe (here is the sample):
df <- structure(list(user_id = c(1L, 1L, 1L, 1L, 1L, 1L), obs_id = c(1L,
2L, 2L, 2L, 2L, 2L), scroll_id = c(3L, 1L, 2L, 3L, 4L, 5L), timestamp = c(-1.74966971796047,
-1.70403832189443, -1.70379906928687, -1.70361867040459, -1.70347088963619,
-1.70319128699835), row_num = 1:6, scroll_length = c(6, 9, 14,
12, 13, 26), x_mean = c(-1.74134749014902, -1.19087086808828,
1.36178725012622, -1.32786301490502, 1.24184201608646, -1.31953110973881
), y_mean = c(-4.93507461932646, 0.0304680987883223, 0.140001980341645,
0.61911843405746, 0.434230282460559, 0.438563278736709), dx_mean = c(-0.514034686928457,
-0.709482080612108, 0.924636289935977, -0.702980646737082, 0.515080876392673,
-0.359676884238743), dy_mean = c(0.972265996197407, -0.692113718739584,
-0.162463490249733, -0.373682612876388, -0.0663766957581004,
0.293619375985922)), .Names = c("user_id", "obs_id", "scroll_id",
"timestamp", "row_num", "scroll_length", "x_mean", "y_mean",
"dx_mean", "dy_mean"), row.names = c(NA, -6L), class = c("tbl_df",
"tbl", "data.frame"))
I want to arrange by timestamp column but I get the following error:
data %>% arrange(timestamp)
data %>% arrange("timestamp")
Error in arrange_impl(.data, dots) : Argument 1 is of unsupported type
matrix
Please advise how to make it work. I know that timestamp is a function and matrix but here it's a column and I "want" dplyr
to "understand" that it's a column.
As @sotos asked:
sessionInfo():
R version 3.4.4 (2018-03-15)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 18.04.1 LTS
Matrix products: default
BLAS: /usr/lib/x86_64-linux-gnu/blas/libblas.so.3.7.1
LAPACK: /usr/lib/x86_64-linux-gnu/lapack/liblapack.so.3.7.1
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C LC_TIME=en_IL.UTF-8 LC_COLLATE=en_US.UTF-8 LC_MONETARY=en_IL.UTF-8
[6] LC_MESSAGES=en_US.UTF-8 LC_PAPER=en_IL.UTF-8 LC_NAME=C LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_IL.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] parallel stats graphics grDevices utils datasets methods base
other attached packages:
[1] bindrcpp_0.2.2 rebus_0.1-3 philentropy_0.2.0 reshape2_1.4.3 broom_0.5.0 dummies_1.5.6 hms_0.4.2
[8] anytime_0.3.1 data.table_1.11.8 bit64_0.9-7 bit_1.1-14 car_3.0-2 carData_3.0-2 caret_6.0-80
[15] lattice_0.20-35 xgboost_0.71.2 doSNOW_1.0.16 snow_0.4-3 doMC_1.3.5 iterators_1.0.10 foreach_1.4.4
[22] randomForest_4.6-14 htmlwidgets_1.3 plotly_4.8.0 jsonlite_1.5 pROC_1.13.0 knitr_1.20 lubridate_1.7.4
[29] MASS_7.3-49 chron_2.3-53 forcats_0.3.0 stringr_1.3.1 dplyr_0.7.7 purrr_0.2.5 readr_1.1.1
[36] tidyr_0.8.2 tibble_1.4.2 ggplot2_3.1.0 tidyverse_1.2.1
loaded via a namespace (and not attached):
[1] nlme_3.1-137 dimRed_0.1.0 httr_1.3.1 tools_3.4.4 backports_1.1.2 R6_2.3.0
[7] rpart_4.1-13 rebus.base_0.0-3 lazyeval_0.2.1 colorspace_1.3-2 nnet_7.3-12 withr_2.1.2
[13] tidyselect_0.2.5 curl_3.1 compiler_3.4.4 cli_1.0.1 rvest_0.3.2 xml2_1.2.0
[19] scales_1.0.0 sfsmisc_1.1-2 DEoptimR_1.0-8 robustbase_0.93-3 RApiDatetime_0.0.4 digest_0.6.18
[25] rebus.unicode_0.0-2 foreign_0.8-70 rio_0.5.10 pkgconfig_2.0.2 htmltools_0.3.6 rlang_0.3.0.1
[31] readxl_1.1.0 ddalpha_1.3.4 rstudioapi_0.8 bindr_0.1.1 zip_1.0.0 ModelMetrics_1.2.0
[37] magrittr_1.5 Matrix_1.2-14 Rcpp_0.12.19 munsell_0.5.0 abind_1.4-5 stringi_1.2.4
[43] yaml_2.2.0 plyr_1.8.4 recipes_0.1.3 grid_3.4.4 pls_2.7-0 crayon_1.3.4
[49] rebus.datetimes_0.0-1 haven_1.1.2 splines_3.4.4 pillar_1.3.0 rebus.numbers_0.0-1 codetools_0.2-15
[55] stats4_3.4.4 CVST_0.2-2 magic_1.5-9 glue_1.3.0 modelr_0.1.2 cellranger_1.1.0
[61] gtable_0.2.0 kernlab_0.9-27 assertthat_0.2.0 DRR_0.0.3 openxlsx_4.1.0 gower_0.1.2
[67] prodlim_2018.04.18 class_7.3-14 survival_2.42-3 viridisLite_0.3.0 geometry_0.3-6 timeDate_3043.102
[73] RcppRoll_0.3.0 lava_1.6.3 ipred_0.9-7
r dataframe dplyr tidyverse
I have the following dataframe (here is the sample):
df <- structure(list(user_id = c(1L, 1L, 1L, 1L, 1L, 1L), obs_id = c(1L,
2L, 2L, 2L, 2L, 2L), scroll_id = c(3L, 1L, 2L, 3L, 4L, 5L), timestamp = c(-1.74966971796047,
-1.70403832189443, -1.70379906928687, -1.70361867040459, -1.70347088963619,
-1.70319128699835), row_num = 1:6, scroll_length = c(6, 9, 14,
12, 13, 26), x_mean = c(-1.74134749014902, -1.19087086808828,
1.36178725012622, -1.32786301490502, 1.24184201608646, -1.31953110973881
), y_mean = c(-4.93507461932646, 0.0304680987883223, 0.140001980341645,
0.61911843405746, 0.434230282460559, 0.438563278736709), dx_mean = c(-0.514034686928457,
-0.709482080612108, 0.924636289935977, -0.702980646737082, 0.515080876392673,
-0.359676884238743), dy_mean = c(0.972265996197407, -0.692113718739584,
-0.162463490249733, -0.373682612876388, -0.0663766957581004,
0.293619375985922)), .Names = c("user_id", "obs_id", "scroll_id",
"timestamp", "row_num", "scroll_length", "x_mean", "y_mean",
"dx_mean", "dy_mean"), row.names = c(NA, -6L), class = c("tbl_df",
"tbl", "data.frame"))
I want to arrange by timestamp column but I get the following error:
data %>% arrange(timestamp)
data %>% arrange("timestamp")
Error in arrange_impl(.data, dots) : Argument 1 is of unsupported type
matrix
Please advise how to make it work. I know that timestamp is a function and matrix but here it's a column and I "want" dplyr
to "understand" that it's a column.
As @sotos asked:
sessionInfo():
R version 3.4.4 (2018-03-15)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 18.04.1 LTS
Matrix products: default
BLAS: /usr/lib/x86_64-linux-gnu/blas/libblas.so.3.7.1
LAPACK: /usr/lib/x86_64-linux-gnu/lapack/liblapack.so.3.7.1
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C LC_TIME=en_IL.UTF-8 LC_COLLATE=en_US.UTF-8 LC_MONETARY=en_IL.UTF-8
[6] LC_MESSAGES=en_US.UTF-8 LC_PAPER=en_IL.UTF-8 LC_NAME=C LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_IL.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] parallel stats graphics grDevices utils datasets methods base
other attached packages:
[1] bindrcpp_0.2.2 rebus_0.1-3 philentropy_0.2.0 reshape2_1.4.3 broom_0.5.0 dummies_1.5.6 hms_0.4.2
[8] anytime_0.3.1 data.table_1.11.8 bit64_0.9-7 bit_1.1-14 car_3.0-2 carData_3.0-2 caret_6.0-80
[15] lattice_0.20-35 xgboost_0.71.2 doSNOW_1.0.16 snow_0.4-3 doMC_1.3.5 iterators_1.0.10 foreach_1.4.4
[22] randomForest_4.6-14 htmlwidgets_1.3 plotly_4.8.0 jsonlite_1.5 pROC_1.13.0 knitr_1.20 lubridate_1.7.4
[29] MASS_7.3-49 chron_2.3-53 forcats_0.3.0 stringr_1.3.1 dplyr_0.7.7 purrr_0.2.5 readr_1.1.1
[36] tidyr_0.8.2 tibble_1.4.2 ggplot2_3.1.0 tidyverse_1.2.1
loaded via a namespace (and not attached):
[1] nlme_3.1-137 dimRed_0.1.0 httr_1.3.1 tools_3.4.4 backports_1.1.2 R6_2.3.0
[7] rpart_4.1-13 rebus.base_0.0-3 lazyeval_0.2.1 colorspace_1.3-2 nnet_7.3-12 withr_2.1.2
[13] tidyselect_0.2.5 curl_3.1 compiler_3.4.4 cli_1.0.1 rvest_0.3.2 xml2_1.2.0
[19] scales_1.0.0 sfsmisc_1.1-2 DEoptimR_1.0-8 robustbase_0.93-3 RApiDatetime_0.0.4 digest_0.6.18
[25] rebus.unicode_0.0-2 foreign_0.8-70 rio_0.5.10 pkgconfig_2.0.2 htmltools_0.3.6 rlang_0.3.0.1
[31] readxl_1.1.0 ddalpha_1.3.4 rstudioapi_0.8 bindr_0.1.1 zip_1.0.0 ModelMetrics_1.2.0
[37] magrittr_1.5 Matrix_1.2-14 Rcpp_0.12.19 munsell_0.5.0 abind_1.4-5 stringi_1.2.4
[43] yaml_2.2.0 plyr_1.8.4 recipes_0.1.3 grid_3.4.4 pls_2.7-0 crayon_1.3.4
[49] rebus.datetimes_0.0-1 haven_1.1.2 splines_3.4.4 pillar_1.3.0 rebus.numbers_0.0-1 codetools_0.2-15
[55] stats4_3.4.4 CVST_0.2-2 magic_1.5-9 glue_1.3.0 modelr_0.1.2 cellranger_1.1.0
[61] gtable_0.2.0 kernlab_0.9-27 assertthat_0.2.0 DRR_0.0.3 openxlsx_4.1.0 gower_0.1.2
[67] prodlim_2018.04.18 class_7.3-14 survival_2.42-3 viridisLite_0.3.0 geometry_0.3-6 timeDate_3043.102
[73] RcppRoll_0.3.0 lava_1.6.3 ipred_0.9-7
r dataframe dplyr tidyverse
r dataframe dplyr tidyverse
edited Nov 8 at 9:19
asked Nov 8 at 9:09
steves
477211
477211
Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Nov 8 at 11:21
add a comment |
Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Nov 8 at 11:21
Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Nov 8 at 11:21
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– Samuel Liew♦
Nov 8 at 11:21
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2 Answers
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2
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df %>% arrange(timestamp)
arrange() from dplyr arranges as below in ascending order
# A tibble: 6 x 10
user_id obs_id scroll_id timestamp row_num scroll_length x_mean y_mean dx_mean dy_mean
<int> <int> <int> <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 3 -1.75 1 6 -1.74 -4.94 -0.514 0.972
2 1 2 1 -1.70 2 9 -1.19 0.0305 -0.709 -0.692
3 1 2 2 -1.70 3 14 1.36 0.140 0.925 -0.162
4 1 2 3 -1.70 4 12 -1.33 0.619 -0.703 -0.374
5 1 2 4 -1.70 5 13 1.24 0.434 0.515 -0.0664
6 1 2 5 -1.70 6 26 -1.32 0.439 -0.360 0.294
add a comment |
up vote
1
down vote
Thanks for all comments, I have found out a solution:
My df
dataframe is scaled and centered - the function that produces df returns:
scale(df)
When I have printed str(df)
I have seen attributes saying that it's centered and scaled.
When converting to data.table
it solved the issue:
df %>% as.data.table() %>% dplyr::arrange(obs_id, user_id, scroll_id, timestamp)
Please correct me if I am wrong.
Nice to see the solution.. also observed uneven number of data points in each variable like dx_mean, dy_mean have only 212 points vs obs_id which is 356 rows?
– Sai Prabhanjan Reddy
Nov 8 at 11:16
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
df %>% arrange(timestamp)
arrange() from dplyr arranges as below in ascending order
# A tibble: 6 x 10
user_id obs_id scroll_id timestamp row_num scroll_length x_mean y_mean dx_mean dy_mean
<int> <int> <int> <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 3 -1.75 1 6 -1.74 -4.94 -0.514 0.972
2 1 2 1 -1.70 2 9 -1.19 0.0305 -0.709 -0.692
3 1 2 2 -1.70 3 14 1.36 0.140 0.925 -0.162
4 1 2 3 -1.70 4 12 -1.33 0.619 -0.703 -0.374
5 1 2 4 -1.70 5 13 1.24 0.434 0.515 -0.0664
6 1 2 5 -1.70 6 26 -1.32 0.439 -0.360 0.294
add a comment |
up vote
2
down vote
df %>% arrange(timestamp)
arrange() from dplyr arranges as below in ascending order
# A tibble: 6 x 10
user_id obs_id scroll_id timestamp row_num scroll_length x_mean y_mean dx_mean dy_mean
<int> <int> <int> <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 3 -1.75 1 6 -1.74 -4.94 -0.514 0.972
2 1 2 1 -1.70 2 9 -1.19 0.0305 -0.709 -0.692
3 1 2 2 -1.70 3 14 1.36 0.140 0.925 -0.162
4 1 2 3 -1.70 4 12 -1.33 0.619 -0.703 -0.374
5 1 2 4 -1.70 5 13 1.24 0.434 0.515 -0.0664
6 1 2 5 -1.70 6 26 -1.32 0.439 -0.360 0.294
add a comment |
up vote
2
down vote
up vote
2
down vote
df %>% arrange(timestamp)
arrange() from dplyr arranges as below in ascending order
# A tibble: 6 x 10
user_id obs_id scroll_id timestamp row_num scroll_length x_mean y_mean dx_mean dy_mean
<int> <int> <int> <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 3 -1.75 1 6 -1.74 -4.94 -0.514 0.972
2 1 2 1 -1.70 2 9 -1.19 0.0305 -0.709 -0.692
3 1 2 2 -1.70 3 14 1.36 0.140 0.925 -0.162
4 1 2 3 -1.70 4 12 -1.33 0.619 -0.703 -0.374
5 1 2 4 -1.70 5 13 1.24 0.434 0.515 -0.0664
6 1 2 5 -1.70 6 26 -1.32 0.439 -0.360 0.294
df %>% arrange(timestamp)
arrange() from dplyr arranges as below in ascending order
# A tibble: 6 x 10
user_id obs_id scroll_id timestamp row_num scroll_length x_mean y_mean dx_mean dy_mean
<int> <int> <int> <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 3 -1.75 1 6 -1.74 -4.94 -0.514 0.972
2 1 2 1 -1.70 2 9 -1.19 0.0305 -0.709 -0.692
3 1 2 2 -1.70 3 14 1.36 0.140 0.925 -0.162
4 1 2 3 -1.70 4 12 -1.33 0.619 -0.703 -0.374
5 1 2 4 -1.70 5 13 1.24 0.434 0.515 -0.0664
6 1 2 5 -1.70 6 26 -1.32 0.439 -0.360 0.294
edited Nov 8 at 9:27
DJV
1,2231216
1,2231216
answered Nov 8 at 9:24
Sai Prabhanjan Reddy
1829
1829
add a comment |
add a comment |
up vote
1
down vote
Thanks for all comments, I have found out a solution:
My df
dataframe is scaled and centered - the function that produces df returns:
scale(df)
When I have printed str(df)
I have seen attributes saying that it's centered and scaled.
When converting to data.table
it solved the issue:
df %>% as.data.table() %>% dplyr::arrange(obs_id, user_id, scroll_id, timestamp)
Please correct me if I am wrong.
Nice to see the solution.. also observed uneven number of data points in each variable like dx_mean, dy_mean have only 212 points vs obs_id which is 356 rows?
– Sai Prabhanjan Reddy
Nov 8 at 11:16
add a comment |
up vote
1
down vote
Thanks for all comments, I have found out a solution:
My df
dataframe is scaled and centered - the function that produces df returns:
scale(df)
When I have printed str(df)
I have seen attributes saying that it's centered and scaled.
When converting to data.table
it solved the issue:
df %>% as.data.table() %>% dplyr::arrange(obs_id, user_id, scroll_id, timestamp)
Please correct me if I am wrong.
Nice to see the solution.. also observed uneven number of data points in each variable like dx_mean, dy_mean have only 212 points vs obs_id which is 356 rows?
– Sai Prabhanjan Reddy
Nov 8 at 11:16
add a comment |
up vote
1
down vote
up vote
1
down vote
Thanks for all comments, I have found out a solution:
My df
dataframe is scaled and centered - the function that produces df returns:
scale(df)
When I have printed str(df)
I have seen attributes saying that it's centered and scaled.
When converting to data.table
it solved the issue:
df %>% as.data.table() %>% dplyr::arrange(obs_id, user_id, scroll_id, timestamp)
Please correct me if I am wrong.
Thanks for all comments, I have found out a solution:
My df
dataframe is scaled and centered - the function that produces df returns:
scale(df)
When I have printed str(df)
I have seen attributes saying that it's centered and scaled.
When converting to data.table
it solved the issue:
df %>% as.data.table() %>% dplyr::arrange(obs_id, user_id, scroll_id, timestamp)
Please correct me if I am wrong.
answered Nov 8 at 10:50
steves
477211
477211
Nice to see the solution.. also observed uneven number of data points in each variable like dx_mean, dy_mean have only 212 points vs obs_id which is 356 rows?
– Sai Prabhanjan Reddy
Nov 8 at 11:16
add a comment |
Nice to see the solution.. also observed uneven number of data points in each variable like dx_mean, dy_mean have only 212 points vs obs_id which is 356 rows?
– Sai Prabhanjan Reddy
Nov 8 at 11:16
Nice to see the solution.. also observed uneven number of data points in each variable like dx_mean, dy_mean have only 212 points vs obs_id which is 356 rows?
– Sai Prabhanjan Reddy
Nov 8 at 11:16
Nice to see the solution.. also observed uneven number of data points in each variable like dx_mean, dy_mean have only 212 points vs obs_id which is 356 rows?
– Sai Prabhanjan Reddy
Nov 8 at 11:16
add a comment |
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Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Nov 8 at 11:21