Data Cleaning in R: remove test customer names - r

Question

I am handling customer data that has customer first and last name. I want to clean the names of any random keystrokes. Test accounts are jumbled in the data-set and have junk names. For example in the below data I want to remove customers 2,5,9,10,12 etc. I would appreciate your help.
Customer Id FirstName LastName
1 MARY MEYER
2 GFRTYUIO UHBVYY
3 CHARLES BEAL
4 MARNI MONTANEZ
5 GDTDTTD DTTHDTHTHTHD
6 TIFFANY BAYLESS
7 CATHRYN JONES
8 TINA CUNNINGHAM
9 FGCYFCGCGFC FGCGFCHGHG
10 ADDHJSDLG DHGAHG
11 WALTER FINN
12 GFCTFCGCFGC CG GFCGFCGFCGF
13 ASDASDASD AASDASDASD
14 TYKTYKYTKTY YTKTYKTYK
15 HFHFHF HAVE
16 REBECCA CROSSWHITE
17 GHSGHG HGASGH
18 JESSICA TREMBLEY
19 GFRTYUIO UHBVYY
20 HUBHGBUHBUH YTVYVFYVYFFV
21 HEATHER WYRICK
22 JASON SPLICHAL
23 RUSTY OWENS
24 DUSTIN WILLIAMS
25 GFCGFCFGCGFC GRCGFXFGDGF
26 QWQWQW QWQWWW
27 LIWNDVLIHWDV LIAENVLIHEAV
28 DARLENE SHORTRIDGE
29 BETH HDHDHDH
30 ROBERT SHIELDS
31 GHERDHBXFH DFHFDHDFH
32 ACE TESSSSSRT
33 ALLISON AWTREY
34 UYGUGVHGVGHVG HGHGVUYYU
35 HCJHV FHJSEFHSIEHF

Answer 1

The problem seems to be that you'd need a solid definition of improbable names, and that is not really related to R. Anyway, I suggest you go by the first names and remove all those names that are not plausible. As a source of plausible first names or positive list, you could use e.g. SSA Baby Name Database. This should work reasonably well to filter out English first names. If you have more location specific needs for first names, just look online for other baby name databases and try to scrape them as a positive list.
Once you have them in a vector named positiveNames, filter out all non-positive names like this:
data_new <- data_original[!data_original$firstName %in% positiveNames,]

Answer 2

My approach is the following:
1) Merge FirstName and LastName into a single string, strname.
Then, count the number of letters for each strname.
2) At this point, we find that for real names, like "MARNIMONTANEZ", are composed of two 'M'; two 'A'; one 'R'; one 'I'; three 'N'; one 'O'; one 'T'.
And we find that fake names, like "GFCTFCGCFGCCGGFCGFCGFCGF", are composed of six 'G'; five 'F'; 8 'C'.
3) The pattern to distinguish real names from fake names becomes clear:
real names are characterized by a more variety of letters. We can measure this by creating a variable check_real computed as: number of unique letters / total string length
fake names are characterized by few letters repeated several times. We can measure this by creating a variable check_fake computed as: average frequency of each letter
4) Finally, we just have to define a threshold to identify an anomaly for both variable. In the cases where these threshold are triggered, a flag_real and a flag_fake appears.
if flag_real == 1 & flag_fake == 0, the name is real
if flag_real == 0 & flag_fake == 1, the name is fake
In the rare cases when the two flags agrees (i.e. flag_real == 1 & flag_fake == 1), you have to investigate the record manually to optimize the threshold.

Answer 3

You can calculate variability strength of full name (combine FirstName and LastName) by calculating length of unique letters in full name divided by total number of characters in the full name. Then, just remove the names that has low variability strength. This means that you are removing the names that has a high frequency of same random keystrokes resulting in low variability strength.
I did this using charToRaw function because it very faster and using dplyr library, as below:
# Building Test Data
df <- data.frame(CustomerId = c(1, 2, 3, 4, 5, 6, 7),
FirstName = c("MARY", "FGCYFCGCGFC", "GFCTFCGCFGC", "ASDASDASD", "GDTDTTD", "WALTER", "GFCTFCGCFGC"),
LastName = c("MEYER", "FGCGFCHGHG", "GFCGFCGFCGF", "AASDASDASD", "DTTHDTHTHTHD", "FINN", "CG GFCGFCGFCGF"), stringsAsFactors = FALSE)
#test data: df
# CustomerId FirstName LastName
#1 1 MARY MEYER
#2 2 FGCYFCGCGFC FGCGFCHGHG
#3 3 GFCTFCGCFGC GFCGFCGFCGF
#4 4 ASDASDASD AASDASDASD
#5 5 GDTDTTD DTTHDTHTHTHD
#6 6 WALTER FINN
#7 7 GFCTFCGCFGC CG GFCGFCGFCGF
library(dplyr)
df %>%
## Combining FirstName and LastName
mutate(FullName = paste(FirstName, gsub(" ", "", LastName, fixed = TRUE))) %>%
group_by(FullName) %>%
## Calculating variability strength for each full name
mutate(Variability = length(unique(as.integer(charToRaw(FullName))))/nchar(FullName))%>%
## Filtering full name, I set above or equal to 0.4 (You can change this)
## Meaning we are keeping full name that has variability strength greater than or equal to 0.40
filter(Variability >= 0.40)
# A tibble: 2 x 5
# Groups: FullName [2]
# CustomerId FirstName LastName FullName Variability
# <dbl> <chr> <chr> <chr> <dbl>
#1 1 MARY MEYER MARY MEYER 0.6000000
#2 6 WALTER FINN WALTER FINN 0.9090909

Answer 4

I tried to combine the suggestions in the below code. Thanks everyone for the help.
# load required libraries
library(hunspell)
library(dplyr)
# read data in dataframe df
df<-data.frame(CustomerId = c(1, 2, 3, 4, 5, 6, 7,8),
FirstName = c("MARY"," ALBERT SAM", "FGCYFCGCGFC", "GFCTFCGCFGC", "ASDASDASD", "GDTDTTD", "WALTER", "GFCTFCGCFGC"),
LastName = c("MEYER","TEST", "FGCGFCHGHG", "GFCGFCGFCGF", "AASDASDASD", "DTTHDTHTHTHD", "FINN", "CG GFCGFCGFCGF"), stringsAsFactors = FALSE)
# Keep unique names
df<-distinct(df,FirstName, LastName, .keep_all = TRUE)
# Spell check using hunspel
df$flag <- hunspell_check(df$FirstName) | hunspell_check(as.character(df$LastName))
# remove middle names
df$FirstNameOnly<-gsub(" .*","",df$FirstName)
# SSA name data using https://www.ssa.gov/oact/babynames/names.zip
# unzip files in folder named names
files<-list.files("/names",pattern="*.txt")
ssa_names<- do.call(rbind, lapply(files, function(x) read.csv(x,
col.names = c("Name","Gender","Frequency"),stringsAsFactors = FALSE)))
# Change SSA names to uppercase
ssa_names$Name <- toupper(ssa_names$Name)
# Flad for SSA names
df$flag_SSA<-ifelse(df$FirstNameOnly %in% ssa_names$Name,TRUE,FALSE)
rm(ssa_names)
# remove spaces and concatenate first name and last name
df$strname<-gsub(" ","",paste(df$FirstName,df$LastName, sep = ""))
# Name string length
df$len<-nchar(df$strname)
# Unique string length
for(n in 1:nrow(df))
{
df$ulen[n]<-length(unique(strsplit(df$strname[n], "")[[1]]))
}
# Ratio variable for unique string length over total string length
df$ratio<-ifelse(df$len==0,0,df$ulen/df$len)
# Histogram to determine cutoff ratio
hist(df$ratio)
test<-df[df$ratio<.4 & df$flag_SSA==FALSE & df$flag==FALSE,]