CODEWITHSUNDEEP
rdataframedata-structuresdata-science
melil
melil

Reputation: 81

Sort the order of rows based on the column names of another dataframe

The dat columnnames corresponds with the values of one of the columns of dat.info, namely dat.info$eset.sample.
I reordered the dat columnames and now want to reorder the rows of dat.info in the same order as the column names in dat.

Before:
dat

|          | 3              | 2           |1          |
| -------- | -------------- |-------------|-----------|
| x        | Order          |This         |In         |
| y        | Way            |Correct      |The        |

dat.info

|        |eset.sample     | 
| -------| -------------- |
| now    | 3              |
| order  | 2              |
| same   | 1              |

After:
dat

|          | 1              | 2           |3          |
| -------- | -------------- |-------------|-----------|
| x        | In             |This         |Order      |
| y        | The            |Correct      |Way        |

dat.info

|        |eset.sample     | 
| -------| -------------- |
| same   | 1              |
| order  | 2              |
| now    | 3              |

> dput(dat[1,1:40])
structure(list(GSM97800 = 4701.5, GSM97803 = 4735, GSM97804 = 2863.9, 
    GSM97805 = 5350.2, GSM97807 = 4789.4, GSM97809 = 5837.8, 
    GSM97811 = 4446.7, GSM97812 = 4264.1, GSM97816 = 11011.5, 
    GSM97817 = 3832.6, GSM97820 = 5227.2, GSM97825 = 2935.6, 
    GSM97827 = 3561.5, GSM97828 = 10728.8, GSM97833 = 4156.6, 
    GSM97834 = 4278.4, GSM97840 = 5916.2, GSM97846 = 3136.6, 
    GSM97848 = 4415.7, GSM97849 = 3313.4, GSM97850 = 6011.4, 
    GSM97853 = 4107.2, GSM97855 = 7053.4, GSM97878 = 8196.2, 
    GSM97913 = 6006.6, GSM97932 = 2619.5, GSM97939 = 11699.2, 
    GSM97951 = 7718.5, GSM97957 = 12700.4, GSM97972 = 8816.4, 
    GSM97793 = 10178.1, GSM97795 = 7826.6, GSM97802 = 6857.8, 
    GSM97810 = 8255, GSM97815 = 8016.8, GSM97837 = 10350.9, GSM97843 = 10626.1, 
    GSM97890 = 8584.5, GSM97899 = 10615.1, GSM97910 = 12047.3), row.names = "1007_s_at", class = "data.frame")

> dput(dat.info[1:40,1])
c("GSM97800", "GSM97803", "GSM97804", "GSM97805", "GSM97807", 
"GSM97809", "GSM97811", "GSM97812", "GSM97816", "GSM97817", "GSM97820", 
"GSM97825", "GSM97827", "GSM97828", "GSM97833", "GSM97834", "GSM97840", 
"GSM97846", "GSM97848", "GSM97849", "GSM97850", "GSM97853", "GSM97855", 
"GSM97878", "GSM97913", "GSM97932", "GSM97939", "GSM97951", "GSM97957", 
"GSM97972", "GSM97793", "GSM97795", "GSM97802", "GSM97810", "GSM97815", 
"GSM97837", "GSM97843", "GSM97890", "GSM97899", "GSM97910")

dat.new <- cbind(dat[1:23], dat[24:30], dat[131:168], dat[31:49], dat[169:180], dat[50:130])

> dput(dat.new[1,1:40])
structure(list(GSM97800 = 4701.5, GSM97803 = 4735, GSM97804 = 2863.9, 
    GSM97805 = 5350.2, GSM97807 = 4789.4, GSM97809 = 5837.8, 
    GSM97811 = 4446.7, GSM97812 = 4264.1, GSM97816 = 11011.5, 
    GSM97817 = 3832.6, GSM97820 = 5227.2, GSM97825 = 2935.6, 
    GSM97827 = 3561.5, GSM97828 = 10728.8, GSM97833 = 4156.6, 
    GSM97834 = 4278.4, GSM97840 = 5916.2, GSM97846 = 3136.6, 
    GSM97848 = 4415.7, GSM97849 = 3313.4, GSM97850 = 6011.4, 
    GSM97853 = 4107.2, GSM97855 = 7053.4, GSM97878 = 8196.2, 
    GSM97913 = 6006.6, GSM97932 = 2619.5, GSM97939 = 11699.2, 
    GSM97951 = 7718.5, GSM97957 = 12700.4, GSM97972 = 8816.4, 
    GSM97799 = 9267.2, GSM97823 = 6581.8, GSM97824 = 11527.5, 
    GSM97830 = 12217, GSM97835 = 10019.1, GSM97838 = 4566.5, 
    GSM97841 = 6601, GSM97842 = 10721.4, GSM97854 = 4918.9, GSM97857 = 10116.9), row.names = "1007_s_at", class = "data.frame")

Upvotes: 0

Views: 959

Answers (2)

Julian_Hn
Julian_Hn

Reputation: 2141

We can use order on the colnames and apply this to the dat.info matrix:

dat <- matrix(c("Order","Way","This","Correct","In","The"),ncol=3)
dimnames(dat) <- list(c("x","y"),c("3","2","1"))

dat.info <- matrix(c(3,2,1),ncol=1)
dimnames(dat.info) <- list(c("now","order","same"),"eset.sample")


dat.new <- dat[,sort(colnames(dat))]
dat.info.new <- dat.info[order(colnames(dat)),,drop=F]

dat.new

  1     2         3      
x "In"  "This"    "Order"
y "The" "Correct" "Way"  

dat.info.new

     eset.sample
same            1
order           2
now             3

Without reproducible example an idea how it could work when your order in dat.new is not based on some form of sort:

ord <- match(colnames(dat),colnames(dat.new))
dat.info.new <- dat.info[ord,,drop=F]

Upvotes: 0

TarJae
TarJae

Reputation: 78917

In case the dat dataframe is at it is: You could try this:

dat %>% 
  mutate(id = row_number()) %>% 
  pivot_longer(
    cols = -id,
    names_to = "name",
    values_to = "value"
  ) %>% 
  group_by(id) %>% 
  arrange(name, .by_group = TRUE) %>% 
  pivot_wider(
    names_from = name
  )

output:

  `1`   `2`     `3`  
  <chr> <chr>   <chr>
1 In    This    Order
2 The   Correct Way

data:

structure(list(`1` = c("In", "The"), `2` = c("This", "Correct"
), `3` = c("Order", "Way")), row.names = c(NA, -2L), class = c("tbl_df", 
"tbl", "data.frame"))

Upvotes: 2

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