如何根据r中元素的第一位保留行？

Question

I have the following data.frame

crime<-c(71040,142320,71013,71013,72113)
coded.month<-c("2018-10","2018-10","2018-10","2018-10","2018-10")
df<-data.frame(coded.month,crime)
  coded.month  crime
1     2018-10  71040
2     2018-10 142320
3     2018-10  71013
4     2018-10  71013
5     2018-10  72113

Bascially, I want to isolate all the rows where the first digit of crime is 7 so that I get the following

  coded.month  crime
1     2018-10  71040
3     2018-10  71013
4     2018-10  71013
5     2018-10  72113

how do I go about this?

Answer 1

You may use substr :

df[substr(df$crime, 0, 1) == 7, ]
#   coded.month crime
# 1     2018-10 71040
# 3     2018-10 71013
# 4     2018-10 71013
# 5     2018-10 72113

Answer 2

We can also use %/%

df[df$crime%/% 10000 == 7, ]
#    coded.month crime
#1     2018-10 71040
#3     2018-10 71013
#4     2018-10 71013
#5     2018-10 72113

Answer 3

using startsWith :

subset(df, startsWith(as.character(crime),"7"))
#   coded.month crime
# 1     2018-10 71040
# 3     2018-10 71013
# 4     2018-10 71013
# 5     2018-10 72113

Answer 4

This also involves converting values (implicitly) to strings, but it works:

 df[grep("^7", df$crime), ]

Edit: a purely numerical solution:

df[floor(df$crime /  10^floor(log10(df$crime))) == 7, ]

Answer 5

By defining a new dataframe using grepl() to match only those df$crime values that start with "7":

df_new <- df[grepl("^7", df$crime, perl = T),]
df_new
  coded.month crime
1     2018-10 71040
3     2018-10 71013
4     2018-10 71013
5     2018-10 72113