How to extract values from an R data frame column that do not start and end with certain characters?



Sometimes we just want to extract the values of a data column based on initial and ending values of a column that has strings or sometimes the values of a column that has strings are recorded with some extra characters and we want to extract those values. For this purpose, we can use negation of grepl with single square brackets.

Example

Consider the below data frame −

> x2<-c("Alabama", "Alaska", "American Samoa", "Arizona", "Arkansas", "California", "Colorado", "Connecticut", "Delaware", "District of Columbia", "Florida", "Georgia", "Guam", "Hawaii", "Idaho", "Illinois", "Indiana", "Iowa", "Kansas", "Kentucky", "Louisiana", "Maine", "Maryland", "Massachusetts", "Michigan", "Minnesota", "Minor Outlying Islands", "Mississippi", "Missouri", "Montana", "Nebraska", "Nevada", "New Hampshire", "New Jersey", "New Mexico", "New York", "North Carolina", "North Dakota", "Northern Mariana Islands", "Ohio", "Oklahoma", "Oregon", "Pennsylvania", "Puerto Rico", "Rhode Island", "South Carolina", "South Dakota", "Tennessee", "Texas", "U.S. Virgin Islands", "Utah", "Vermont", "Virginia", "Washington", "West Virginia", "Wisconsin", "Wyoming")
> df2<-data.frame(x2) > head(df2,20)

Output

x2 1 Alabama 2 Alaska 3 American Samoa 4 Arizona 5 Arkansas 6 California 7 Colorado 8 Connecticut 9 Delaware 10 District of Columbia 11 Florida 12 Georgia 13 Guam 14 Hawaii 15 Idaho 16 Illinois 17 Indiana 18 Iowa 19 Kansas 20 Kentucky

Finding states that neither start with A nor ends with a −

> df2[!grepl("^A|a$",df2$x2),]

Output

[1] Colorado Connecticut Delaware [4] Guam Hawaii Idaho [7] Illinois Kansas Kentucky [10] Maine Maryland Massachusetts [13] Michigan Minor Outlying Islands Mississippi [16] Missouri New Hampshire New Jersey [19] New Mexico New York Northern Mariana Islands [22] Ohio Oregon Puerto Rico [25] Rhode Island Tennessee Texas [28] U.S. Virgin Islands Utah Vermont [31] Washington Wisconsin Wyoming 57 Levels: Alabama Alaska American Samoa Arizona Arkansas ... Wyoming

Let’s have a look at another example −

> x1<- c("Indiaaa","Chinaaa","Russiaa","Canadaaa","Indonesiaaa","Croatiaaa","Mauritaniaaa"," Albaniaaa","Angolaaa","Armeniaaa","Malaysiaaa","Maltaaa","Boliviaaa","Burmaaa","Pa nama","Romaniaa","Saudi-Arabia","Serbiaaa","Syriaaa","Tongaaa","Koreaaa","Libya")
> y1<-sample(1:10,22,replace=TRUE) > df1<-data.frame(x1,y1) > df1

Output

x1 y1 1 Indiaaa 6 2 Chinaaa 1 3 Russiaa 9 4 Canadaaa 7 5 Indonesiaaa 7 6 Croatiaaa 3 7 Mauritaniaaa 6 8 Albaniaaa 2 9 Angolaaa 10 10 Armeniaaa 10 11 Malaysiaaa 7 12 Maltaaa 3 13 Boliviaaa 2 14 Burmaaa 10 15 Panama 1 16 Romaniaa 10 17 Saudi-Arabia 10 18 Serbiaaa 8 19 Syriaaa 10 20 Tongaaa 5 21 Koreaaa 7 22 Libya 8
> df1[!grepl("^A|aa$",df1$x1),]

Output

x1 y1 15 Panama 1 17 Saudi-Arabia 10 22 Libya 8
> df1[!grepl("^S|aa$",df1$x1),]

Output

x1 y1 15 Panama 1 22 Libya 8
> df1[!grepl("^B|aa$",df1$x1),]

Output

x1 y1 15 Panama 1 17 Saudi-Arabia 10 22 Libya 8
> df1[!grepl("^P|aa$",df1$x1),]

Output

x1 y1 17 Saudi-Arabia 10 22 Libya 8
> df1[!grepl("^L|aa$",df1$x1),]

Output

x1 y1 15 Panama 1 17 Saudi-Arabia 10
Updated on: 2020-09-07T06:23:01+05:30

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