deleting rows without filtering other values
1
Say I have a dataframe like this:
NAME YEAR PERCENTAGE
A 2001 NA
A 2002 NA
A 2003 5.0
B 2001 3.3
B 2002 2.3
B 2003 NA
I want remove rows with NA by selecting the specific rows:
NAME YEAR PERCENTAGE
A 2003 5.0
B 2001 3.3
B 2002 2.3
and then change B to A,expected output like this:
NAME YEAR PERCENTAGE
A 2001 3.3
A 2002 2.3
A 2003 5.0
I tried subset(),but since I have other values, it would filter other values which should be remained.
r dplyr subset
edited Nov 29 '18 at 1:02
Ronak Shah
44.8k104266
asked Nov 29 '18 at 0:28
wawawawawawa
587
add a comment |
1
Say I have a dataframe like this:
NAME YEAR PERCENTAGE
A 2001 NA
A 2002 NA
A 2003 5.0
B 2001 3.3
B 2002 2.3
B 2003 NA
I want remove rows with NA by selecting the specific rows:
NAME YEAR PERCENTAGE
A 2003 5.0
B 2001 3.3
B 2002 2.3
and then change B to A,expected output like this:
NAME YEAR PERCENTAGE
A 2001 3.3
A 2002 2.3
A 2003 5.0
I tried subset(),but since I have other values, it would filter other values which should be remained.
r dplyr subset
edited Nov 29 '18 at 1:02
Ronak Shah
44.8k104266
asked Nov 29 '18 at 0:28
wawawawawawa
587
3
What aboutdf[!is.na(df$PERCENTAGE),]?
– CIAndrews
Nov 29 '18 at 0:34
1
Do you need to order the output byYEAR? You don't mention it, but this is the case in your expected output
– prosoitos
Nov 29 '18 at 0:50
ordering by name and then order the year of each name
– wawawa
Nov 29 '18 at 0:52
add a comment |
1
1
1
Say I have a dataframe like this:
NAME YEAR PERCENTAGE
A 2001 NA
A 2002 NA
A 2003 5.0
B 2001 3.3
B 2002 2.3
B 2003 NA
I want remove rows with NA by selecting the specific rows:
NAME YEAR PERCENTAGE
A 2003 5.0
B 2001 3.3
B 2002 2.3
and then change B to A,expected output like this:
NAME YEAR PERCENTAGE
A 2001 3.3
A 2002 2.3
A 2003 5.0
I tried subset(),but since I have other values, it would filter other values which should be remained.
r dplyr subset
edited Nov 29 '18 at 1:02
Ronak Shah
44.8k104266
asked Nov 29 '18 at 0:28
wawawawawawa
587
Say I have a dataframe like this:
NAME YEAR PERCENTAGE
A 2001 NA
A 2002 NA
A 2003 5.0
B 2001 3.3
B 2002 2.3
B 2003 NA
I want remove rows with NA by selecting the specific rows:
NAME YEAR PERCENTAGE
A 2003 5.0
B 2001 3.3
B 2002 2.3
and then change B to A,expected output like this:
NAME YEAR PERCENTAGE
A 2001 3.3
A 2002 2.3
A 2003 5.0
I tried subset(),but since I have other values, it would filter other values which should be remained.
r dplyr subset
r dplyr subset
edited Nov 29 '18 at 1:02
Ronak Shah
44.8k104266
asked Nov 29 '18 at 0:28
wawawawawawa
587
edited Nov 29 '18 at 1:02
Ronak Shah
44.8k104266
asked Nov 29 '18 at 0:28
wawawawawawa
587
edited Nov 29 '18 at 1:02
Ronak Shah
44.8k104266
edited Nov 29 '18 at 1:02
Ronak Shah
44.8k104266
edited Nov 29 '18 at 1:02
Ronak Shah
44.8k104266
44.8k104266
asked Nov 29 '18 at 0:28
wawawawawawa
587
asked Nov 29 '18 at 0:28
wawawawawawa
587
asked Nov 29 '18 at 0:28
wawawawawawa
587
587
3
What aboutdf[!is.na(df$PERCENTAGE),]?
– CIAndrews
Nov 29 '18 at 0:34
1
Do you need to order the output byYEAR? You don't mention it, but this is the case in your expected output
– prosoitos
Nov 29 '18 at 0:50
ordering by name and then order the year of each name
– wawawa
Nov 29 '18 at 0:52
add a comment |
3
What aboutdf[!is.na(df$PERCENTAGE),]?
– CIAndrews
Nov 29 '18 at 0:34
1
Do you need to order the output byYEAR? You don't mention it, but this is the case in your expected output
– prosoitos
Nov 29 '18 at 0:50
ordering by name and then order the year of each name
– wawawa
Nov 29 '18 at 0:52
3
3
What about
df[!is.na(df$PERCENTAGE),] ?– CIAndrews
Nov 29 '18 at 0:34
What about
df[!is.na(df$PERCENTAGE),] ?– CIAndrews
Nov 29 '18 at 0:34
1
1
Do you need to order the output by
YEAR? You don't mention it, but this is the case in your expected output– prosoitos
Nov 29 '18 at 0:50
Do you need to order the output by
YEAR? You don't mention it, but this is the case in your expected output– prosoitos
Nov 29 '18 at 0:50
ordering by name and then order the year of each name
– wawawa
Nov 29 '18 at 0:52
ordering by name and then order the year of each name
– wawawa
Nov 29 '18 at 0:52
add a comment |
2 Answers
2
active
oldest
votes
4
With the library dplyr, you have access to several functions (like filter(), arrange() or mutate()) that able you to modify you dataframe:
# the dataframe
df <- data.frame(
NAME = rep(c('A', 'B'), each = 3),
YEAR = rep(2001:2003, length = 6),
PERC = c(NA, NA, 5, 3.3, 2.3, NA)
)
# load the library
library(dplyr)
df %>%
filter(!is.na(PERC)) %>% # filter missing values
arrange(YEAR) %>% # order according YEAR
mutate(NAME = replace(NAME, NAME == 'B', 'A')) # change values
# result
NAME YEAR PERC
1 A 2001 3.3
2 A 2002 2.3
3 A 2003 5.0
answered Nov 29 '18 at 0:39
demarsylvaindemarsylvain
744214
1
These are alldplyrfunctions, so justlibrary(dplyr)is enough.
– neilfws
Nov 29 '18 at 0:41
you're right. I will ajust the post.
– demarsylvain
Nov 29 '18 at 0:45
add a comment |
1
Assuming your data frame is called df:
library(dplyr)
df %>% na.omit() %>% mutate(NAME = "A")
Result:
NAME YEAR PERC
1 A 2003 5.0
2 A 2001 3.3
3 A 2002 2.3
answered Nov 29 '18 at 0:47
prosoitosprosoitos
935520
1
This works for the example data, but it would (a) omit rows that are NA in any column and (b) change any name to "A", not just "B" to "A".
– neilfws
Nov 29 '18 at 0:50
1
It is unclear in the question whether those are requirements or not
– prosoitos
Nov 29 '18 at 0:51
Note that I agree with you :) and the other answer is much more general (+1). But I went with the minimal code given the info in the question
– prosoitos
Nov 29 '18 at 0:55
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
With the library dplyr, you have access to several functions (like filter(), arrange() or mutate()) that able you to modify you dataframe:
# the dataframe
df <- data.frame(
NAME = rep(c('A', 'B'), each = 3),
YEAR = rep(2001:2003, length = 6),
PERC = c(NA, NA, 5, 3.3, 2.3, NA)
)
# load the library
library(dplyr)
df %>%
filter(!is.na(PERC)) %>% # filter missing values
arrange(YEAR) %>% # order according YEAR
mutate(NAME = replace(NAME, NAME == 'B', 'A')) # change values
# result
NAME YEAR PERC
1 A 2001 3.3
2 A 2002 2.3
3 A 2003 5.0
answered Nov 29 '18 at 0:39
demarsylvaindemarsylvain
744214
1
These are alldplyrfunctions, so justlibrary(dplyr)is enough.
– neilfws
Nov 29 '18 at 0:41
you're right. I will ajust the post.
– demarsylvain
Nov 29 '18 at 0:45
add a comment |
4
With the library dplyr, you have access to several functions (like filter(), arrange() or mutate()) that able you to modify you dataframe:
# the dataframe
df <- data.frame(
NAME = rep(c('A', 'B'), each = 3),
YEAR = rep(2001:2003, length = 6),
PERC = c(NA, NA, 5, 3.3, 2.3, NA)
)
# load the library
library(dplyr)
df %>%
filter(!is.na(PERC)) %>% # filter missing values
arrange(YEAR) %>% # order according YEAR
mutate(NAME = replace(NAME, NAME == 'B', 'A')) # change values
# result
NAME YEAR PERC
1 A 2001 3.3
2 A 2002 2.3
3 A 2003 5.0
answered Nov 29 '18 at 0:39
demarsylvaindemarsylvain
744214
1
These are alldplyrfunctions, so justlibrary(dplyr)is enough.
– neilfws
Nov 29 '18 at 0:41
you're right. I will ajust the post.
– demarsylvain
Nov 29 '18 at 0:45
add a comment |
4
4
4
With the library dplyr, you have access to several functions (like filter(), arrange() or mutate()) that able you to modify you dataframe:
# the dataframe
df <- data.frame(
NAME = rep(c('A', 'B'), each = 3),
YEAR = rep(2001:2003, length = 6),
PERC = c(NA, NA, 5, 3.3, 2.3, NA)
)
# load the library
library(dplyr)
df %>%
filter(!is.na(PERC)) %>% # filter missing values
arrange(YEAR) %>% # order according YEAR
mutate(NAME = replace(NAME, NAME == 'B', 'A')) # change values
# result
NAME YEAR PERC
1 A 2001 3.3
2 A 2002 2.3
3 A 2003 5.0
answered Nov 29 '18 at 0:39
demarsylvaindemarsylvain
744214
With the library dplyr, you have access to several functions (like filter(), arrange() or mutate()) that able you to modify you dataframe:
# the dataframe
df <- data.frame(
NAME = rep(c('A', 'B'), each = 3),
YEAR = rep(2001:2003, length = 6),
PERC = c(NA, NA, 5, 3.3, 2.3, NA)
)
# load the library
library(dplyr)
df %>%
filter(!is.na(PERC)) %>% # filter missing values
arrange(YEAR) %>% # order according YEAR
mutate(NAME = replace(NAME, NAME == 'B', 'A')) # change values
# result
NAME YEAR PERC
1 A 2001 3.3
2 A 2002 2.3
3 A 2003 5.0
answered Nov 29 '18 at 0:39
demarsylvaindemarsylvain
744214
edited Nov 29 '18 at 0:46
answered Nov 29 '18 at 0:39
demarsylvaindemarsylvain
744214
answered Nov 29 '18 at 0:39
demarsylvaindemarsylvain
744214
answered Nov 29 '18 at 0:39
demarsylvaindemarsylvain
744214
744214
1
These are alldplyrfunctions, so justlibrary(dplyr)is enough.
– neilfws
Nov 29 '18 at 0:41
you're right. I will ajust the post.
– demarsylvain
Nov 29 '18 at 0:45
add a comment |
1
These are alldplyrfunctions, so justlibrary(dplyr)is enough.
– neilfws
Nov 29 '18 at 0:41
you're right. I will ajust the post.
– demarsylvain
Nov 29 '18 at 0:45
1
1
These are all
dplyr functions, so just library(dplyr) is enough.– neilfws
Nov 29 '18 at 0:41
These are all
dplyr functions, so just library(dplyr) is enough.– neilfws
Nov 29 '18 at 0:41
you're right. I will ajust the post.
– demarsylvain
Nov 29 '18 at 0:45
you're right. I will ajust the post.
– demarsylvain
Nov 29 '18 at 0:45
add a comment |
1
Assuming your data frame is called df:
library(dplyr)
df %>% na.omit() %>% mutate(NAME = "A")
Result:
NAME YEAR PERC
1 A 2003 5.0
2 A 2001 3.3
3 A 2002 2.3
answered Nov 29 '18 at 0:47
prosoitosprosoitos
935520
1
This works for the example data, but it would (a) omit rows that are NA in any column and (b) change any name to "A", not just "B" to "A".
– neilfws
Nov 29 '18 at 0:50
1
It is unclear in the question whether those are requirements or not
– prosoitos
Nov 29 '18 at 0:51
Note that I agree with you :) and the other answer is much more general (+1). But I went with the minimal code given the info in the question
– prosoitos
Nov 29 '18 at 0:55
add a comment |
1
Assuming your data frame is called df:
library(dplyr)
df %>% na.omit() %>% mutate(NAME = "A")
Result:
NAME YEAR PERC
1 A 2003 5.0
2 A 2001 3.3
3 A 2002 2.3
answered Nov 29 '18 at 0:47
prosoitosprosoitos
935520
1
This works for the example data, but it would (a) omit rows that are NA in any column and (b) change any name to "A", not just "B" to "A".
– neilfws
Nov 29 '18 at 0:50
1
It is unclear in the question whether those are requirements or not
– prosoitos
Nov 29 '18 at 0:51
Note that I agree with you :) and the other answer is much more general (+1). But I went with the minimal code given the info in the question
– prosoitos
Nov 29 '18 at 0:55
add a comment |
1
1
1
Assuming your data frame is called df:
library(dplyr)
df %>% na.omit() %>% mutate(NAME = "A")
Result:
NAME YEAR PERC
1 A 2003 5.0
2 A 2001 3.3
3 A 2002 2.3
answered Nov 29 '18 at 0:47
prosoitosprosoitos
935520
Assuming your data frame is called df:
library(dplyr)
df %>% na.omit() %>% mutate(NAME = "A")
Result:
NAME YEAR PERC
1 A 2003 5.0
2 A 2001 3.3
3 A 2002 2.3
answered Nov 29 '18 at 0:47
prosoitosprosoitos
935520
answered Nov 29 '18 at 0:47
prosoitosprosoitos
935520
answered Nov 29 '18 at 0:47
prosoitosprosoitos
935520
answered Nov 29 '18 at 0:47
prosoitosprosoitos
935520
935520
1
This works for the example data, but it would (a) omit rows that are NA in any column and (b) change any name to "A", not just "B" to "A".
– neilfws
Nov 29 '18 at 0:50
1
It is unclear in the question whether those are requirements or not
– prosoitos
Nov 29 '18 at 0:51
Note that I agree with you :) and the other answer is much more general (+1). But I went with the minimal code given the info in the question
– prosoitos
Nov 29 '18 at 0:55
add a comment |
1
This works for the example data, but it would (a) omit rows that are NA in any column and (b) change any name to "A", not just "B" to "A".
– neilfws
Nov 29 '18 at 0:50
1
It is unclear in the question whether those are requirements or not
– prosoitos
Nov 29 '18 at 0:51
Note that I agree with you :) and the other answer is much more general (+1). But I went with the minimal code given the info in the question
– prosoitos
Nov 29 '18 at 0:55
1
1
This works for the example data, but it would (a) omit rows that are NA in any column and (b) change any name to "A", not just "B" to "A".
– neilfws
Nov 29 '18 at 0:50
This works for the example data, but it would (a) omit rows that are NA in any column and (b) change any name to "A", not just "B" to "A".
– neilfws
Nov 29 '18 at 0:50
1
1
It is unclear in the question whether those are requirements or not
– prosoitos
Nov 29 '18 at 0:51
It is unclear in the question whether those are requirements or not
– prosoitos
Nov 29 '18 at 0:51
Note that I agree with you :) and the other answer is much more general (+1). But I went with the minimal code given the info in the question
– prosoitos
Nov 29 '18 at 0:55
Note that I agree with you :) and the other answer is much more general (+1). But I went with the minimal code given the info in the question
– prosoitos
Nov 29 '18 at 0:55
add a comment |
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3
What about
df[!is.na(df$PERCENTAGE),]?– CIAndrews
Nov 29 '18 at 0:34
1
Do you need to order the output by
YEAR? You don't mention it, but this is the case in your expected output– prosoitos
Nov 29 '18 at 0:50
ordering by name and then order the year of each name
– wawawa
Nov 29 '18 at 0:52