Order dataframe by column of another dataframe
Consider the following df: df1 <-data.frame('GID'=c('GID1','GID3','GID2','GID1','GID2'), 'Sequence'= c(4,7,6,2,3)) df2 <- data.frame('GID' = c('GID3','GID1','GID2','GID1','GID2'),'Trial'=c('SA1','SA5','ES4','ES3','ES9'))
I want to order df1 by the column df2$GID such that I could cbind the df2$Trial column into df1. I know match can do that but match only finds the first occurrence of values and I have repeated values. Thus I'm confused on the way to do that as this is a subset of a large data.frame.
r
asked Nov 23 '18 at 16:37
Alexandre Mondaini
986
add a comment |
Consider the following df: df1 <-data.frame('GID'=c('GID1','GID3','GID2','GID1','GID2'), 'Sequence'= c(4,7,6,2,3)) df2 <- data.frame('GID' = c('GID3','GID1','GID2','GID1','GID2'),'Trial'=c('SA1','SA5','ES4','ES3','ES9'))
I want to order df1 by the column df2$GID such that I could cbind the df2$Trial column into df1. I know match can do that but match only finds the first occurrence of values and I have repeated values. Thus I'm confused on the way to do that as this is a subset of a large data.frame.
r
asked Nov 23 '18 at 16:37
Alexandre Mondaini
986
You have duplicates in both datasets for column 'GID' May need to create a sequence column
– akrun
Nov 23 '18 at 16:39
create a sequence column for df2 and then ?
– Alexandre Mondaini
Nov 23 '18 at 16:46
add a comment |
Consider the following df: df1 <-data.frame('GID'=c('GID1','GID3','GID2','GID1','GID2'), 'Sequence'= c(4,7,6,2,3)) df2 <- data.frame('GID' = c('GID3','GID1','GID2','GID1','GID2'),'Trial'=c('SA1','SA5','ES4','ES3','ES9'))
I want to order df1 by the column df2$GID such that I could cbind the df2$Trial column into df1. I know match can do that but match only finds the first occurrence of values and I have repeated values. Thus I'm confused on the way to do that as this is a subset of a large data.frame.
r
asked Nov 23 '18 at 16:37
Alexandre Mondaini
986
Consider the following df: df1 <-data.frame('GID'=c('GID1','GID3','GID2','GID1','GID2'), 'Sequence'= c(4,7,6,2,3)) df2 <- data.frame('GID' = c('GID3','GID1','GID2','GID1','GID2'),'Trial'=c('SA1','SA5','ES4','ES3','ES9'))
I want to order df1 by the column df2$GID such that I could cbind the df2$Trial column into df1. I know match can do that but match only finds the first occurrence of values and I have repeated values. Thus I'm confused on the way to do that as this is a subset of a large data.frame.
r
r
asked Nov 23 '18 at 16:37
Alexandre Mondaini
986
asked Nov 23 '18 at 16:37
Alexandre Mondaini
986
asked Nov 23 '18 at 16:37
Alexandre Mondaini
986
asked Nov 23 '18 at 16:37
Alexandre Mondaini
986
asked Nov 23 '18 at 16:37
Alexandre Mondaini
986
986
You have duplicates in both datasets for column 'GID' May need to create a sequence column
– akrun
Nov 23 '18 at 16:39
create a sequence column for df2 and then ?
– Alexandre Mondaini
Nov 23 '18 at 16:46
add a comment |
You have duplicates in both datasets for column 'GID' May need to create a sequence column
– akrun
Nov 23 '18 at 16:39
create a sequence column for df2 and then ?
– Alexandre Mondaini
Nov 23 '18 at 16:46
You have duplicates in both datasets for column 'GID' May need to create a sequence column
– akrun
Nov 23 '18 at 16:39
You have duplicates in both datasets for column 'GID' May need to create a sequence column
– akrun
Nov 23 '18 at 16:39
create a sequence column for df2 and then ?
– Alexandre Mondaini
Nov 23 '18 at 16:46
create a sequence column for df2 and then ?
– Alexandre Mondaini
Nov 23 '18 at 16:46
add a comment |
1 Answer
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Have you tried using match this way: df1 <- df1[match(df2$GID, df1$GID),] (make sure to review the results)?
GID Sequence
2 GID3 7
1 GID1 4
3 GID2 6
1.1 GID1 4
3.1 GID2 6
Do you want to get the cartesian product of GID, Sequence and Trial?
Also, could there be any GID of df2 that is not present in df1 and how would you like to handle that?
Do you have any data acting as a "key", identifying uniquely each entry?
answered Nov 23 '18 at 16:47
LostIT
8318
No, I don't need the cartesian product.
– Alexandre Mondaini
Nov 23 '18 at 17:19
No, I don't need the cartesian product. the df1 was first shorter than df2. df1 had no repeated values while df2 had. Then I expanded df1 as followsdf1[df2$GID,]. Thus all GID from df2 are present now in df1 and replicated by df2. What identifies uniquely each GID is the combination of the two columns of df2 e.g:df2$GID,df2$Trial. Such combination is unique and is in fact what I want to transfer to df1.
– Alexandre Mondaini
Nov 23 '18 at 17:25
Actually,df1, does not have each value of GID matched to a unique value of Sequence, thus I can't understand how it doesn't have or shouldn't have replicates.
– LostIT
Nov 23 '18 at 17:44
actually is the GID column of df2 that has a unique Trial. So df2 dataframe has duplicated GIDs but unique Trial numbers for the duplicated GIDs . Do you understand what I mean ?
– Alexandre Mondaini
Nov 23 '18 at 18:00
This can't be true, since e.g. GID1 is matched to both SA5 and ES3 trials. Do you mean that the "key" of df2, is the Trial and a single Trial value can only be matched to one GID value?
– LostIT
Nov 23 '18 at 18:06
|
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Have you tried using match this way: df1 <- df1[match(df2$GID, df1$GID),] (make sure to review the results)?
GID Sequence
2 GID3 7
1 GID1 4
3 GID2 6
1.1 GID1 4
3.1 GID2 6
Do you want to get the cartesian product of GID, Sequence and Trial?
Also, could there be any GID of df2 that is not present in df1 and how would you like to handle that?
Do you have any data acting as a "key", identifying uniquely each entry?
answered Nov 23 '18 at 16:47
LostIT
8318
No, I don't need the cartesian product.
– Alexandre Mondaini
Nov 23 '18 at 17:19
No, I don't need the cartesian product. the df1 was first shorter than df2. df1 had no repeated values while df2 had. Then I expanded df1 as followsdf1[df2$GID,]. Thus all GID from df2 are present now in df1 and replicated by df2. What identifies uniquely each GID is the combination of the two columns of df2 e.g:df2$GID,df2$Trial. Such combination is unique and is in fact what I want to transfer to df1.
– Alexandre Mondaini
Nov 23 '18 at 17:25
Actually,df1, does not have each value of GID matched to a unique value of Sequence, thus I can't understand how it doesn't have or shouldn't have replicates.
– LostIT
Nov 23 '18 at 17:44
actually is the GID column of df2 that has a unique Trial. So df2 dataframe has duplicated GIDs but unique Trial numbers for the duplicated GIDs . Do you understand what I mean ?
– Alexandre Mondaini
Nov 23 '18 at 18:00
This can't be true, since e.g. GID1 is matched to both SA5 and ES3 trials. Do you mean that the "key" of df2, is the Trial and a single Trial value can only be matched to one GID value?
– LostIT
Nov 23 '18 at 18:06
|
show 1 more comment
Have you tried using match this way: df1 <- df1[match(df2$GID, df1$GID),] (make sure to review the results)?
GID Sequence
2 GID3 7
1 GID1 4
3 GID2 6
1.1 GID1 4
3.1 GID2 6
Do you want to get the cartesian product of GID, Sequence and Trial?
Also, could there be any GID of df2 that is not present in df1 and how would you like to handle that?
Do you have any data acting as a "key", identifying uniquely each entry?
answered Nov 23 '18 at 16:47
LostIT
8318
No, I don't need the cartesian product.
– Alexandre Mondaini
Nov 23 '18 at 17:19
No, I don't need the cartesian product. the df1 was first shorter than df2. df1 had no repeated values while df2 had. Then I expanded df1 as followsdf1[df2$GID,]. Thus all GID from df2 are present now in df1 and replicated by df2. What identifies uniquely each GID is the combination of the two columns of df2 e.g:df2$GID,df2$Trial. Such combination is unique and is in fact what I want to transfer to df1.
– Alexandre Mondaini
Nov 23 '18 at 17:25
Actually,df1, does not have each value of GID matched to a unique value of Sequence, thus I can't understand how it doesn't have or shouldn't have replicates.
– LostIT
Nov 23 '18 at 17:44
actually is the GID column of df2 that has a unique Trial. So df2 dataframe has duplicated GIDs but unique Trial numbers for the duplicated GIDs . Do you understand what I mean ?
– Alexandre Mondaini
Nov 23 '18 at 18:00
This can't be true, since e.g. GID1 is matched to both SA5 and ES3 trials. Do you mean that the "key" of df2, is the Trial and a single Trial value can only be matched to one GID value?
– LostIT
Nov 23 '18 at 18:06
|
show 1 more comment
Have you tried using match this way: df1 <- df1[match(df2$GID, df1$GID),] (make sure to review the results)?
GID Sequence
2 GID3 7
1 GID1 4
3 GID2 6
1.1 GID1 4
3.1 GID2 6
Do you want to get the cartesian product of GID, Sequence and Trial?
Also, could there be any GID of df2 that is not present in df1 and how would you like to handle that?
Do you have any data acting as a "key", identifying uniquely each entry?
answered Nov 23 '18 at 16:47
LostIT
8318
Have you tried using match this way: df1 <- df1[match(df2$GID, df1$GID),] (make sure to review the results)?
GID Sequence
2 GID3 7
1 GID1 4
3 GID2 6
1.1 GID1 4
3.1 GID2 6
Do you want to get the cartesian product of GID, Sequence and Trial?
Also, could there be any GID of df2 that is not present in df1 and how would you like to handle that?
Do you have any data acting as a "key", identifying uniquely each entry?
answered Nov 23 '18 at 16:47
LostIT
8318
edited Nov 23 '18 at 17:01
answered Nov 23 '18 at 16:47
LostIT
8318
answered Nov 23 '18 at 16:47
LostIT
8318
answered Nov 23 '18 at 16:47
LostIT
8318
8318
No, I don't need the cartesian product.
– Alexandre Mondaini
Nov 23 '18 at 17:19
No, I don't need the cartesian product. the df1 was first shorter than df2. df1 had no repeated values while df2 had. Then I expanded df1 as followsdf1[df2$GID,]. Thus all GID from df2 are present now in df1 and replicated by df2. What identifies uniquely each GID is the combination of the two columns of df2 e.g:df2$GID,df2$Trial. Such combination is unique and is in fact what I want to transfer to df1.
– Alexandre Mondaini
Nov 23 '18 at 17:25
Actually,df1, does not have each value of GID matched to a unique value of Sequence, thus I can't understand how it doesn't have or shouldn't have replicates.
– LostIT
Nov 23 '18 at 17:44
actually is the GID column of df2 that has a unique Trial. So df2 dataframe has duplicated GIDs but unique Trial numbers for the duplicated GIDs . Do you understand what I mean ?
– Alexandre Mondaini
Nov 23 '18 at 18:00
This can't be true, since e.g. GID1 is matched to both SA5 and ES3 trials. Do you mean that the "key" of df2, is the Trial and a single Trial value can only be matched to one GID value?
– LostIT
Nov 23 '18 at 18:06
|
show 1 more comment
No, I don't need the cartesian product.
– Alexandre Mondaini
Nov 23 '18 at 17:19
No, I don't need the cartesian product. the df1 was first shorter than df2. df1 had no repeated values while df2 had. Then I expanded df1 as followsdf1[df2$GID,]. Thus all GID from df2 are present now in df1 and replicated by df2. What identifies uniquely each GID is the combination of the two columns of df2 e.g:df2$GID,df2$Trial. Such combination is unique and is in fact what I want to transfer to df1.
– Alexandre Mondaini
Nov 23 '18 at 17:25
Actually,df1, does not have each value of GID matched to a unique value of Sequence, thus I can't understand how it doesn't have or shouldn't have replicates.
– LostIT
Nov 23 '18 at 17:44
actually is the GID column of df2 that has a unique Trial. So df2 dataframe has duplicated GIDs but unique Trial numbers for the duplicated GIDs . Do you understand what I mean ?
– Alexandre Mondaini
Nov 23 '18 at 18:00
This can't be true, since e.g. GID1 is matched to both SA5 and ES3 trials. Do you mean that the "key" of df2, is the Trial and a single Trial value can only be matched to one GID value?
– LostIT
Nov 23 '18 at 18:06
No, I don't need the cartesian product.
– Alexandre Mondaini
Nov 23 '18 at 17:19
No, I don't need the cartesian product.
– Alexandre Mondaini
Nov 23 '18 at 17:19
No, I don't need the cartesian product. the df1 was first shorter than df2. df1 had no repeated values while df2 had. Then I expanded df1 as follows
df1[df2$GID,]. Thus all GID from df2 are present now in df1 and replicated by df2. What identifies uniquely each GID is the combination of the two columns of df2 e.g: df2$GID,df2$Trial. Such combination is unique and is in fact what I want to transfer to df1.– Alexandre Mondaini
Nov 23 '18 at 17:25
No, I don't need the cartesian product. the df1 was first shorter than df2. df1 had no repeated values while df2 had. Then I expanded df1 as follows
df1[df2$GID,]. Thus all GID from df2 are present now in df1 and replicated by df2. What identifies uniquely each GID is the combination of the two columns of df2 e.g: df2$GID,df2$Trial. Such combination is unique and is in fact what I want to transfer to df1.– Alexandre Mondaini
Nov 23 '18 at 17:25
Actually,
df1, does not have each value of GID matched to a unique value of Sequence, thus I can't understand how it doesn't have or shouldn't have replicates.– LostIT
Nov 23 '18 at 17:44
Actually,
df1, does not have each value of GID matched to a unique value of Sequence, thus I can't understand how it doesn't have or shouldn't have replicates.– LostIT
Nov 23 '18 at 17:44
actually is the GID column of df2 that has a unique Trial. So df2 dataframe has duplicated GIDs but unique Trial numbers for the duplicated GIDs . Do you understand what I mean ?
– Alexandre Mondaini
Nov 23 '18 at 18:00
actually is the GID column of df2 that has a unique Trial. So df2 dataframe has duplicated GIDs but unique Trial numbers for the duplicated GIDs . Do you understand what I mean ?
– Alexandre Mondaini
Nov 23 '18 at 18:00
This can't be true, since e.g. GID1 is matched to both SA5 and ES3 trials. Do you mean that the "key" of df2, is the Trial and a single Trial value can only be matched to one GID value?
– LostIT
Nov 23 '18 at 18:06
This can't be true, since e.g. GID1 is matched to both SA5 and ES3 trials. Do you mean that the "key" of df2, is the Trial and a single Trial value can only be matched to one GID value?
– LostIT
Nov 23 '18 at 18:06
|
show 1 more comment
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You have duplicates in both datasets for column 'GID' May need to create a sequence column
– akrun
Nov 23 '18 at 16:39
create a sequence column for df2 and then ?
– Alexandre Mondaini
Nov 23 '18 at 16:46