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How to separate a dataframe based on specific string in column name [duplicate]

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1

This question already has an answer here:

• 
  Split string by last two characters in R? (/negative string indices)
  
  5 answers
  
  

I have a huge data that I cannot split into two sets

 df<- structure(list(name = structure(1:3, .Label = c("a", "b", "c"
), class = "factor"), X3C_AALI_01A = c(651L, 2L, 1877L), X3C_AALJ_01B = c(419L, 
2L, 1825L), X3C_AALK_01A = c(1310L, 52L, 1286L), X4H_AAAK_11B = c(2978L, 
4L, 1389L), X5L_AAT0_01B = c(2576L, 15L, 1441L), X5L_AAT1_01A = c(2886L, 
5L, 921L), X5T_A9QA_03A = c(929L, 3L, 935L), A1_A0SI_10A = c(1578L, 
1L, 2217L), A1_A0SK_07C = c(3003L, 6L, 2984L), A1_A0SO_01A = c(6413L, 
0L, 3577L), A1_A0SP_05B = c(5157L, 5L, 4596L), A2_A04P_01A = c(4283L, 
6L, 2508L), X5L_AAh1_10A = c(2886L, 5L, 921L), X5T_A0QA_03A = c(929L, 
3L, 935L), A1_A0Sm_10A = c(1578L, 1L, 2217L), A1_ArSK_01A = c(3003L, 
6L, 2984L), A1_AfSO_01A = c(6413L, 0L, 3577L), A1_AuSP_05A = c(5157L, 
5L, 4596L), A2_Ap4P_11A = c(4283L, 6L, 2508L)), class = "data.frame", row.names = c(NA, 
-3L))

basically , I want to split the data based on the last character of the column name. for example if you look at the above data, the second column is like this 3C_AALI_01A which I want to generate two data sets based on the _01A

So those columns that have 01 to 09 values I want them to be in one data frame and those ones that have 10 to whatever number want them to be in the second data frame. For example in the above example data.

the columns with the following names should be in one data frame

3C_AALI_01A
3C_AALJ_01B
3C_AALK_01A
5L_AAT0_01B
5L_AAT1_01A
5T_A9QA_03A
A1_A0SK_07C
A1_A0SO_01A
A1_A0SP_05B
A2_A04P_01A
5T_A0QA_03A
A1_ArSK_01A
A1_AfSO_01A
A1_AuSP_05A

and the columns with the following names should be in another data frame

4H_AAAK_11B
A1_A0SI_10A
5L_AAh1_10A
A1_A0Sm_10A
A2_Ap4P_11A

r

share|improve this question

edited Apr 10 at 10:45

smci

15.8k679110

asked Mar 22 at 22:13

LearnerLearner

38010

marked as duplicate by smci, Maurits Evers r
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Apr 12 at 5:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Really you want negative string indexing. Also, your dataframe is kind of transposed, it would be more normal to have one single column name with the names, and numerical columns a, b, c. Like t(df) without the unwanted coercion to string. If your dataframe was structured like that, you could use tidyr::separate(..., sep=-1)
  
  – smci
  Apr 10 at 10:47
  

  
  
  
  

add a comment | 

1

This question already has an answer here:

• 
  Split string by last two characters in R? (/negative string indices)
  
  5 answers
  
  

I have a huge data that I cannot split into two sets

 df<- structure(list(name = structure(1:3, .Label = c("a", "b", "c"
), class = "factor"), X3C_AALI_01A = c(651L, 2L, 1877L), X3C_AALJ_01B = c(419L, 
2L, 1825L), X3C_AALK_01A = c(1310L, 52L, 1286L), X4H_AAAK_11B = c(2978L, 
4L, 1389L), X5L_AAT0_01B = c(2576L, 15L, 1441L), X5L_AAT1_01A = c(2886L, 
5L, 921L), X5T_A9QA_03A = c(929L, 3L, 935L), A1_A0SI_10A = c(1578L, 
1L, 2217L), A1_A0SK_07C = c(3003L, 6L, 2984L), A1_A0SO_01A = c(6413L, 
0L, 3577L), A1_A0SP_05B = c(5157L, 5L, 4596L), A2_A04P_01A = c(4283L, 
6L, 2508L), X5L_AAh1_10A = c(2886L, 5L, 921L), X5T_A0QA_03A = c(929L, 
3L, 935L), A1_A0Sm_10A = c(1578L, 1L, 2217L), A1_ArSK_01A = c(3003L, 
6L, 2984L), A1_AfSO_01A = c(6413L, 0L, 3577L), A1_AuSP_05A = c(5157L, 
5L, 4596L), A2_Ap4P_11A = c(4283L, 6L, 2508L)), class = "data.frame", row.names = c(NA, 
-3L))

basically , I want to split the data based on the last character of the column name. for example if you look at the above data, the second column is like this 3C_AALI_01A which I want to generate two data sets based on the _01A

So those columns that have 01 to 09 values I want them to be in one data frame and those ones that have 10 to whatever number want them to be in the second data frame. For example in the above example data.

the columns with the following names should be in one data frame

3C_AALI_01A
3C_AALJ_01B
3C_AALK_01A
5L_AAT0_01B
5L_AAT1_01A
5T_A9QA_03A
A1_A0SK_07C
A1_A0SO_01A
A1_A0SP_05B
A2_A04P_01A
5T_A0QA_03A
A1_ArSK_01A
A1_AfSO_01A
A1_AuSP_05A

and the columns with the following names should be in another data frame

4H_AAAK_11B
A1_A0SI_10A
5L_AAh1_10A
A1_A0Sm_10A
A2_Ap4P_11A

r

share|improve this question

edited Apr 10 at 10:45

smci

15.8k679110

asked Mar 22 at 22:13

LearnerLearner

38010

marked as duplicate by smci, Maurits Evers r
Users with the  r badge can single-handedly close r questions as duplicates and reopen them as needed.

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Apr 12 at 5:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Really you want negative string indexing. Also, your dataframe is kind of transposed, it would be more normal to have one single column name with the names, and numerical columns a, b, c. Like t(df) without the unwanted coercion to string. If your dataframe was structured like that, you could use tidyr::separate(..., sep=-1)
  
  – smci
  Apr 10 at 10:47
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  Split string by last two characters in R? (/negative string indices)
  
  5 answers
  
  

I have a huge data that I cannot split into two sets

 df<- structure(list(name = structure(1:3, .Label = c("a", "b", "c"
), class = "factor"), X3C_AALI_01A = c(651L, 2L, 1877L), X3C_AALJ_01B = c(419L, 
2L, 1825L), X3C_AALK_01A = c(1310L, 52L, 1286L), X4H_AAAK_11B = c(2978L, 
4L, 1389L), X5L_AAT0_01B = c(2576L, 15L, 1441L), X5L_AAT1_01A = c(2886L, 
5L, 921L), X5T_A9QA_03A = c(929L, 3L, 935L), A1_A0SI_10A = c(1578L, 
1L, 2217L), A1_A0SK_07C = c(3003L, 6L, 2984L), A1_A0SO_01A = c(6413L, 
0L, 3577L), A1_A0SP_05B = c(5157L, 5L, 4596L), A2_A04P_01A = c(4283L, 
6L, 2508L), X5L_AAh1_10A = c(2886L, 5L, 921L), X5T_A0QA_03A = c(929L, 
3L, 935L), A1_A0Sm_10A = c(1578L, 1L, 2217L), A1_ArSK_01A = c(3003L, 
6L, 2984L), A1_AfSO_01A = c(6413L, 0L, 3577L), A1_AuSP_05A = c(5157L, 
5L, 4596L), A2_Ap4P_11A = c(4283L, 6L, 2508L)), class = "data.frame", row.names = c(NA, 
-3L))

basically , I want to split the data based on the last character of the column name. for example if you look at the above data, the second column is like this 3C_AALI_01A which I want to generate two data sets based on the _01A

So those columns that have 01 to 09 values I want them to be in one data frame and those ones that have 10 to whatever number want them to be in the second data frame. For example in the above example data.

the columns with the following names should be in one data frame

3C_AALI_01A
3C_AALJ_01B
3C_AALK_01A
5L_AAT0_01B
5L_AAT1_01A
5T_A9QA_03A
A1_A0SK_07C
A1_A0SO_01A
A1_A0SP_05B
A2_A04P_01A
5T_A0QA_03A
A1_ArSK_01A
A1_AfSO_01A
A1_AuSP_05A

and the columns with the following names should be in another data frame

4H_AAAK_11B
A1_A0SI_10A
5L_AAh1_10A
A1_A0Sm_10A
A2_Ap4P_11A

r

share|improve this question

edited Apr 10 at 10:45

smci

15.8k679110

asked Mar 22 at 22:13

LearnerLearner

38010

 df<- structure(list(name = structure(1:3, .Label = c("a", "b", "c"
), class = "factor"), X3C_AALI_01A = c(651L, 2L, 1877L), X3C_AALJ_01B = c(419L, 
2L, 1825L), X3C_AALK_01A = c(1310L, 52L, 1286L), X4H_AAAK_11B = c(2978L, 
4L, 1389L), X5L_AAT0_01B = c(2576L, 15L, 1441L), X5L_AAT1_01A = c(2886L, 
5L, 921L), X5T_A9QA_03A = c(929L, 3L, 935L), A1_A0SI_10A = c(1578L, 
1L, 2217L), A1_A0SK_07C = c(3003L, 6L, 2984L), A1_A0SO_01A = c(6413L, 
0L, 3577L), A1_A0SP_05B = c(5157L, 5L, 4596L), A2_A04P_01A = c(4283L, 
6L, 2508L), X5L_AAh1_10A = c(2886L, 5L, 921L), X5T_A0QA_03A = c(929L, 
3L, 935L), A1_A0Sm_10A = c(1578L, 1L, 2217L), A1_ArSK_01A = c(3003L, 
6L, 2984L), A1_AfSO_01A = c(6413L, 0L, 3577L), A1_AuSP_05A = c(5157L, 
5L, 4596L), A2_Ap4P_11A = c(4283L, 6L, 2508L)), class = "data.frame", row.names = c(NA, 
-3L))

3C_AALI_01A
3C_AALJ_01B
3C_AALK_01A
5L_AAT0_01B
5L_AAT1_01A
5T_A9QA_03A
A1_A0SK_07C
A1_A0SO_01A
A1_A0SP_05B
A2_A04P_01A
5T_A0QA_03A
A1_ArSK_01A
A1_AfSO_01A
A1_AuSP_05A

4H_AAAK_11B
A1_A0SI_10A
5L_AAh1_10A
A1_A0Sm_10A
A2_Ap4P_11A

share|improve this question

edited Apr 10 at 10:45

smci

15.8k679110

asked Mar 22 at 22:13

LearnerLearner

38010

share|improve this question

edited Apr 10 at 10:45

smci

15.8k679110

asked Mar 22 at 22:13

LearnerLearner

38010

edited Apr 10 at 10:45

smci

15.8k679110

edited Apr 10 at 10:45

smci

15.8k679110

asked Mar 22 at 22:13

LearnerLearner

38010

asked Mar 22 at 22:13

LearnerLearner

38010

2 Answers
2

active

oldest

votes

1

df1 <- df[,grep('0[1-9].$',colnames(df))]
df2 <- df[,-grep('0[1-9].$',colnames(df))]

share|improve this answer

edited Mar 22 at 22:48

answered Mar 22 at 22:17

WilWil

1,4371217

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Quotes in at least two of your answers are incorrect. Please edit.
  
  – Roman Luštrik
  Mar 22 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Could you elaborate? Or edit and I will accept?
  
  – Wil
  Mar 22 at 22:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  It was an issue with the keyboard on this device. Do they appear correct to you now?
  
  – Wil
  Mar 22 at 22:49
  

  
  
  
  

add a comment | 

0

You could use tidyr::separate(..., last=-1) approach

• which uses negative string indexing, which is what you really want here


• also, your dataframe is transposed, it would be more normal to have one single column name with the names, and numerical columns a, b, c. Like t(df) without the unwanted coercion to string.

share|improve this answer

answered Apr 10 at 10:42

smcismci

15.8k679110

add a comment | 

1

df1 <- df[,grep('0[1-9].$',colnames(df))]
df2 <- df[,-grep('0[1-9].$',colnames(df))]

share|improve this answer

edited Mar 22 at 22:48

answered Mar 22 at 22:17

WilWil

1,4371217

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Quotes in at least two of your answers are incorrect. Please edit.
  
  – Roman Luštrik
  Mar 22 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Could you elaborate? Or edit and I will accept?
  
  – Wil
  Mar 22 at 22:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  It was an issue with the keyboard on this device. Do they appear correct to you now?
  
  – Wil
  Mar 22 at 22:49
  

  
  
  
  

add a comment | 

1

df1 <- df[,grep('0[1-9].$',colnames(df))]
df2 <- df[,-grep('0[1-9].$',colnames(df))]

share|improve this answer

edited Mar 22 at 22:48

answered Mar 22 at 22:17

WilWil

1,4371217

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Quotes in at least two of your answers are incorrect. Please edit.
  
  – Roman Luštrik
  Mar 22 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Could you elaborate? Or edit and I will accept?
  
  – Wil
  Mar 22 at 22:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  It was an issue with the keyboard on this device. Do they appear correct to you now?
  
  – Wil
  Mar 22 at 22:49
  

  
  
  
  

add a comment | 

df1 <- df[,grep('0[1-9].$',colnames(df))]
df2 <- df[,-grep('0[1-9].$',colnames(df))]

share|improve this answer

edited Mar 22 at 22:48

answered Mar 22 at 22:17

WilWil

1,4371217

df1 <- df[,grep('0[1-9].$',colnames(df))]
df2 <- df[,-grep('0[1-9].$',colnames(df))]

share|improve this answer

edited Mar 22 at 22:48

answered Mar 22 at 22:17

WilWil

1,4371217

answered Mar 22 at 22:17

WilWil

1,4371217

answered Mar 22 at 22:17

WilWil

1,4371217

0

You could use tidyr::separate(..., last=-1) approach

• which uses negative string indexing, which is what you really want here


• also, your dataframe is transposed, it would be more normal to have one single column name with the names, and numerical columns a, b, c. Like t(df) without the unwanted coercion to string.

share|improve this answer

answered Apr 10 at 10:42

smcismci

15.8k679110

add a comment | 

0

You could use tidyr::separate(..., last=-1) approach

• which uses negative string indexing, which is what you really want here


• also, your dataframe is transposed, it would be more normal to have one single column name with the names, and numerical columns a, b, c. Like t(df) without the unwanted coercion to string.

share|improve this answer

answered Apr 10 at 10:42

smcismci

15.8k679110

add a comment | 

You could use tidyr::separate(..., last=-1) approach

• which uses negative string indexing, which is what you really want here


• also, your dataframe is transposed, it would be more normal to have one single column name with the names, and numerical columns a, b, c. Like t(df) without the unwanted coercion to string.

share|improve this answer

answered Apr 10 at 10:42

smcismci

15.8k679110

share|improve this answer

answered Apr 10 at 10:42

smcismci

15.8k679110

answered Apr 10 at 10:42

smcismci

15.8k679110

answered Apr 10 at 10:42

smcismci

15.8k679110