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Asked: 2026-05-30T17:32:12+00:00

I’m trying to select rows in a dataframe where the string contained in a

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I’m trying to select rows in a dataframe where the string contained in a column matches either a regular expression or a substring:

dataframe:

aName   bName   pName   call  alleles   logRatio    strength
AX-11086564 F08_ADN103  2011-02-10_R10  AB  CG  0.363371    10.184215
AX-11086564 A01_CD1919  2011-02-24_R11  BB  GG  -1.352707   9.54909
AX-11086564 B05_CD2920  2011-01-27_R6   AB  CG  -0.183802   9.766334
AX-11086564 D04_CD5950  2011-02-09_R9   AB  CG  0.162586    10.165051
AX-11086564 D07_CD6025  2011-02-10_R10  AB  CG  -0.397097   9.940238
AX-11086564 B05_CD3630  2011-02-02_R7   AA  CC  2.349906    9.153076
AX-11086564 D04_ADN103  2011-02-10_R2   BB  GG  -1.898088   9.872966
AX-11086564 A01_CD2588  2011-01-27_R5   BB  GG  -1.208094   9.239801

For example, I want a dataframe containing only rows that contain ADN in column bName. Secondarily, I would like all rows that contain ADN in column bName and that match 2011-02-10_R2 in column pName.

I tried using functions grep(), agrep() and more but without success…

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1 Answer

  1. Editorial Team
    subset(dat, grepl("ADN", bName)  &  pName == "2011-02-10_R2" )

    Note & (and not && which is not vectorized) and that == (and not = which is assignment).

    Note that you could have used:

     dat[ with(dat,  grepl("ADN", bName)  &  pName == "2011-02-10_R2" ) , ]

    … and that might be preferable when used inside functions, however, that will return NA values for any lines where dat$pName is NA. That defect (which some regard as a feature) could be removed by the addition of & !is.na(dat$pName) to the logical expression.

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