subsetting a data frame that has rows with values in more than one column in R
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0
Entering edit mode
9 weeks ago
pramach1 ▴ 30

I have a large data frame that as several rows and columns. I need to subset the data frame to rows that has values in more than column. This is the data frame.

  1. Sample Typhi Kentucky | 8:i:z6
  2. F675BNARV 0 2000(3%)
  3. F685NARV 0 0
  4. F722NARV 2038340 (9.24%) 2882679 (13.07%)

I want to subset the row number 4 (F722NARV), since it has values in more than one column. How do i do that. I have tried various forms of subset and sapply. Any help regarding this is appreciated.

dataframe R subsets • 302 views
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3
Entering edit mode
9 weeks ago

It's unclear how your data.frame is formatted exactly (you can share part of it via dput(head(df))), but generally speaking the code will look something like this.

library("dplyr")

# If you want all columns except the first to not equal 0.
df |>
  rowwise() |>
  filter(if_all(!1, \(x) x != 0)) |>
  ungroup()

# If you just want more than one column (except the first) to not equal 0.
df |>
  rowwise() |>
  filter(sum(c_across(!1) != 0) > 1) |>
  ungroup()

Again, this code may not work for you depending on how your data.frame is actually formatted, so edit the code as needed or include a reproducible example.

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# If you just want more than one column to not equal 0.
df |>  rowwise() |> filter(sum(c_across(!1) != 0) > 1) |> ungroup()

Worked for what I was looking for. Thank you.

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0
Entering edit mode

Thank you. Will try this. In the meantime here is the dput(head(df))

> dput(head(df2))
structure(list(Sample = c("F675BNARV", "F685NARV", "F715NARV", 
"F717NARV", "F722NARV", "F762NARV"), I.48.z4.z24.1.5...48.z4.z24.1.5 = c("0", 
"0", "0", "0", "2038340 (9.24%)", "0"), Kentucky...8.i.z6 = c("0", 
"0", "0", "0", "2882679 (13.07%)", "0"), Molade.or.Wippra...8.z10.z6 = c("831691 (3.69%)", 
"0", "0", "0", "0", "0"), Montevideo...7.g.m.s.NA = c("0", "530046 (2.21%)", 
"7823859 (39.08%)", "0", "0", "0"), Newport...8.e.h.1.2 = c("0", 
"0", "0", "6689807 (22.29%)", "0", "2864791 (9.66%)")), row.names = c(NA, 
6L), class = "data.frame")
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