creating a new dataframe based on differen data in rows
I'm moving an operation from Excel Power Query to R, which is much faster. The result is I have a data frame with thousands of rows, however, I'm looking to create a sample data frame that includes one row for every different option (factor level)for columns 5:10 of 15 columns, so people can manually test every option (like a truth table?)
I could manually do this, but I wondered if I could do it automatically.
col1 col2 col3
name option1 option2
name2 option1 option2
name3 option1 option2
name4 option2 option1
would be converted into a data frame like this:
col1 col2 col3
name option1 option2
name4 option2 option1
any help would be greatly appreciated.
Chris
r dataframe rstudio tidyverse
asked Nov 14 '18 at 13:42
ChrisChris
9011
add a comment |
I'm moving an operation from Excel Power Query to R, which is much faster. The result is I have a data frame with thousands of rows, however, I'm looking to create a sample data frame that includes one row for every different option (factor level)for columns 5:10 of 15 columns, so people can manually test every option (like a truth table?)
I could manually do this, but I wondered if I could do it automatically.
col1 col2 col3
name option1 option2
name2 option1 option2
name3 option1 option2
name4 option2 option1
would be converted into a data frame like this:
col1 col2 col3
name option1 option2
name4 option2 option1
any help would be greatly appreciated.
Chris
r dataframe rstudio tidyverse
asked Nov 14 '18 at 13:42
ChrisChris
9011
see?duplicated
– Bastien
Nov 14 '18 at 13:46
add a comment |
I'm moving an operation from Excel Power Query to R, which is much faster. The result is I have a data frame with thousands of rows, however, I'm looking to create a sample data frame that includes one row for every different option (factor level)for columns 5:10 of 15 columns, so people can manually test every option (like a truth table?)
I could manually do this, but I wondered if I could do it automatically.
col1 col2 col3
name option1 option2
name2 option1 option2
name3 option1 option2
name4 option2 option1
would be converted into a data frame like this:
col1 col2 col3
name option1 option2
name4 option2 option1
any help would be greatly appreciated.
Chris
r dataframe rstudio tidyverse
asked Nov 14 '18 at 13:42
ChrisChris
9011
I'm moving an operation from Excel Power Query to R, which is much faster. The result is I have a data frame with thousands of rows, however, I'm looking to create a sample data frame that includes one row for every different option (factor level)for columns 5:10 of 15 columns, so people can manually test every option (like a truth table?)
I could manually do this, but I wondered if I could do it automatically.
col1 col2 col3
name option1 option2
name2 option1 option2
name3 option1 option2
name4 option2 option1
would be converted into a data frame like this:
col1 col2 col3
name option1 option2
name4 option2 option1
any help would be greatly appreciated.
Chris
r dataframe rstudio tidyverse
r dataframe rstudio tidyverse
asked Nov 14 '18 at 13:42
ChrisChris
9011
asked Nov 14 '18 at 13:42
ChrisChris
9011
asked Nov 14 '18 at 13:42
ChrisChris
9011
asked Nov 14 '18 at 13:42
ChrisChris
9011
asked Nov 14 '18 at 13:42
ChrisChris
9011
9011
see?duplicated
– Bastien
Nov 14 '18 at 13:46
add a comment |
see?duplicated
– Bastien
Nov 14 '18 at 13:46
see
?duplicated– Bastien
Nov 14 '18 at 13:46
see
?duplicated– Bastien
Nov 14 '18 at 13:46
add a comment |
1 Answer
1
active
oldest
votes
With dplyr:
library(dplyr)
d %>% distinct(col2, col3, .keep_all=T)
# col1 col2 col3
# 1 name option1 option2
# 2 name4 option2 option1
If you want to use distinct only for a subset of columns, you can match first a regex:
d %>%
select(matches("[5-10]|[1]")) %>% # this selects only rows from 5 to 10 or 1 in the name
distinct(.keep_all=T)
This will have your first row "col1", and all the rows "col5" to "col10".
answered Nov 14 '18 at 13:45
RLaveRLave
4,42711023
Thanks for the quick reply. In my data frame I end up with 19 thousand rows still (from 39,000) I'm after a table with about 100 - 200 lines, but perhaps my maths is way off. rather than have a new row for every distinct option, have rows which may take care of many options per row, so reducing the number of lines...does that make any sense?
– Chris
Nov 14 '18 at 14:09
mm no, I'm sorry but you need to be more clear. Try updating your question with a reproducible example where you show what you expect.
– RLave
Nov 14 '18 at 14:15
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
With dplyr:
library(dplyr)
d %>% distinct(col2, col3, .keep_all=T)
# col1 col2 col3
# 1 name option1 option2
# 2 name4 option2 option1
If you want to use distinct only for a subset of columns, you can match first a regex:
d %>%
select(matches("[5-10]|[1]")) %>% # this selects only rows from 5 to 10 or 1 in the name
distinct(.keep_all=T)
This will have your first row "col1", and all the rows "col5" to "col10".
answered Nov 14 '18 at 13:45
RLaveRLave
4,42711023
Thanks for the quick reply. In my data frame I end up with 19 thousand rows still (from 39,000) I'm after a table with about 100 - 200 lines, but perhaps my maths is way off. rather than have a new row for every distinct option, have rows which may take care of many options per row, so reducing the number of lines...does that make any sense?
– Chris
Nov 14 '18 at 14:09
mm no, I'm sorry but you need to be more clear. Try updating your question with a reproducible example where you show what you expect.
– RLave
Nov 14 '18 at 14:15
add a comment |
With dplyr:
library(dplyr)
d %>% distinct(col2, col3, .keep_all=T)
# col1 col2 col3
# 1 name option1 option2
# 2 name4 option2 option1
If you want to use distinct only for a subset of columns, you can match first a regex:
d %>%
select(matches("[5-10]|[1]")) %>% # this selects only rows from 5 to 10 or 1 in the name
distinct(.keep_all=T)
This will have your first row "col1", and all the rows "col5" to "col10".
answered Nov 14 '18 at 13:45
RLaveRLave
4,42711023
Thanks for the quick reply. In my data frame I end up with 19 thousand rows still (from 39,000) I'm after a table with about 100 - 200 lines, but perhaps my maths is way off. rather than have a new row for every distinct option, have rows which may take care of many options per row, so reducing the number of lines...does that make any sense?
– Chris
Nov 14 '18 at 14:09
mm no, I'm sorry but you need to be more clear. Try updating your question with a reproducible example where you show what you expect.
– RLave
Nov 14 '18 at 14:15
add a comment |
With dplyr:
library(dplyr)
d %>% distinct(col2, col3, .keep_all=T)
# col1 col2 col3
# 1 name option1 option2
# 2 name4 option2 option1
If you want to use distinct only for a subset of columns, you can match first a regex:
d %>%
select(matches("[5-10]|[1]")) %>% # this selects only rows from 5 to 10 or 1 in the name
distinct(.keep_all=T)
This will have your first row "col1", and all the rows "col5" to "col10".
answered Nov 14 '18 at 13:45
RLaveRLave
4,42711023
With dplyr:
library(dplyr)
d %>% distinct(col2, col3, .keep_all=T)
# col1 col2 col3
# 1 name option1 option2
# 2 name4 option2 option1
If you want to use distinct only for a subset of columns, you can match first a regex:
d %>%
select(matches("[5-10]|[1]")) %>% # this selects only rows from 5 to 10 or 1 in the name
distinct(.keep_all=T)
This will have your first row "col1", and all the rows "col5" to "col10".
answered Nov 14 '18 at 13:45
RLaveRLave
4,42711023
edited Nov 14 '18 at 14:05
answered Nov 14 '18 at 13:45
RLaveRLave
4,42711023
answered Nov 14 '18 at 13:45
RLaveRLave
4,42711023
answered Nov 14 '18 at 13:45
RLaveRLave
4,42711023
4,42711023
Thanks for the quick reply. In my data frame I end up with 19 thousand rows still (from 39,000) I'm after a table with about 100 - 200 lines, but perhaps my maths is way off. rather than have a new row for every distinct option, have rows which may take care of many options per row, so reducing the number of lines...does that make any sense?
– Chris
Nov 14 '18 at 14:09
mm no, I'm sorry but you need to be more clear. Try updating your question with a reproducible example where you show what you expect.
– RLave
Nov 14 '18 at 14:15
add a comment |
Thanks for the quick reply. In my data frame I end up with 19 thousand rows still (from 39,000) I'm after a table with about 100 - 200 lines, but perhaps my maths is way off. rather than have a new row for every distinct option, have rows which may take care of many options per row, so reducing the number of lines...does that make any sense?
– Chris
Nov 14 '18 at 14:09
mm no, I'm sorry but you need to be more clear. Try updating your question with a reproducible example where you show what you expect.
– RLave
Nov 14 '18 at 14:15
Thanks for the quick reply. In my data frame I end up with 19 thousand rows still (from 39,000) I'm after a table with about 100 - 200 lines, but perhaps my maths is way off. rather than have a new row for every distinct option, have rows which may take care of many options per row, so reducing the number of lines...does that make any sense?
– Chris
Nov 14 '18 at 14:09
Thanks for the quick reply. In my data frame I end up with 19 thousand rows still (from 39,000) I'm after a table with about 100 - 200 lines, but perhaps my maths is way off. rather than have a new row for every distinct option, have rows which may take care of many options per row, so reducing the number of lines...does that make any sense?
– Chris
Nov 14 '18 at 14:09
mm no, I'm sorry but you need to be more clear. Try updating your question with a reproducible example where you show what you expect.
– RLave
Nov 14 '18 at 14:15
mm no, I'm sorry but you need to be more clear. Try updating your question with a reproducible example where you show what you expect.
– RLave
Nov 14 '18 at 14:15
add a comment |
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see
?duplicated– Bastien
Nov 14 '18 at 13:46