我有以下数据。
Company
1 Progressive Corp.
2 Travelers Companies Inc.
3 Progressive Finance Corp.
4 Zurich Insurance Group (3)
5 Zurich Financial Services Ltd
6 Zurich Ltd
7 Berkshire Hathaway Inc.
8 Auto-Owners Insurance Co.
9 Berkshire Finance Inc.
10 AmTrust Financial Services
例如:我需要将所有包含“Zurich”(4,5,6)的字符串替换为“Zurich [some word]”
问题有点模糊,因为我们没有被告知存储数据的结构。但是假设数据存储在数据框中,以下代码应该工作:
company.data = data.frame(
c("Progressive Corp.",
"Travelers Companies Inc.",
"Progressive Finance Corp.",
"Zurich Insurance Group (3)",
"Zurich Financial Services Ltd",
"Zurich Ltd",
"Berkshire Hathaway Inc.",
"Auto-Owners Insurance Co.",
"Berkshire Finance Inc.",
"AmTrust Financial Services"), stringsAsFactors = F)
names(company.data) = "company"
change.index = which(grepl("Zurich", company.data$company))
company.data$company[change.index] = "Zurich"
> company.data
company
1 Progressive Corp.
2 Travelers Companies Inc.
3 Progressive Finance Corp.
4 Zurich
5 Zurich
6 Zurich
7 Berkshire Hathaway Inc.
8 Auto-Owners Insurance Co.
9 Berkshire Finance Inc.
10 AmTrust Financial Services