Reputation: 19
I have a column as bellow:
a)1902-2019: -1.05 mm/yr
b)1902-1961: -0.79 mm/yr
c)1962-2019: -1.43 mm/yr
I want to remove character to be as flow:
a)1902-2019
b)1902-1961
c)1962-2019
Would you mind to help me? Thanks in advance
Upvotes: 0
Views: 43
Reputation: 388982
Another way would be to extract everything until a colon (:
)
sub('(.*):.*', '\\1', df$V1)
#[1] "1902-2019" "1902-1961" "1962-2019"
Upvotes: 0
Reputation: 72919
You may use substr
.
df1 <- transform(df1, V3=substr(V1, 1, 9))
df1
# V1 V2 V3
# 1 1902-2019: -1.05 mm/yr 1.1088551 1902-2019
# 2 1902-1961: -0.79 mm/yr -0.1593216 1902-1961
# 3 1962-2019: -1.43 mm/yr 0.5811273 1962-2019
Data:
df1 <- structure(list(V1 = c("1902-2019: -1.05 mm/yr", "1902-1961: -0.79 mm/yr",
"1962-2019: -1.43 mm/yr"), V2 = c(0.0958290168309483, 2.00653802357232,
0.961576240476421)), class = "data.frame", row.names = c(NA,
-3L))
Upvotes: 1