英文:
Remove special characters in rows in a Data Frame
问题
我有一个包含名为“indicator”的数据框，其中包含混合数字和字母的内容，实际观察值之前有混合的数字和字母。在下面的DF中，混合的字母和数字是6.2-，S1.1，S3.1-和I1.1。
实际的DF：
indicator <- c("6.2- Total number customers per month (average)","S1.1 Total of unique users served per month", "S3.1- Volume of merchandise sold per month", "I1.1 Quantity of bags received per month.")
amount <- c(12, 45, 44, 67)
DF <- data.frame(indicator, amount)
如何使用stringr或regex删除混合的数字和字母？
英文:
I have a dataframe with a row called indicator that have numbers mixed with letters before the actual observation (content) i.e. in the below DF, the mixed letters and numbers are
6.2-, S1.1, S3.1- & I1.1
Actual DF
indicator <- c("6.2- Total number customers per month (average)","S1.1 Total of unique users served per month", "S3.1- Volume of merchandise sold per month",
"I1.1 Quantity of bags received per month.")
amount <- c(12, 45, 44, 67)
DF <- data.frame(indicator, amount)
> DF
indicator amount
1 6.2- Total number customers per month (average) 12
2 S1.1 Total of unique users served per month 45
3 S3.1- Volume of merchandise sold per month 44
4 I1.1 Quantity of bags received per month. 67
How do I remove the mixed numbers and letters using either stringr or regex?
答案1
得分: 2
You could use sub() as follows:
<!-- language: r -->
DF$indicator <- sub("^[A-Z]*\\d(?:\\.\\d+)*-?\\s+", "", DF$indicator)
DF
正则表达式的解释如下：
^从指标的开头开始匹配[A-Z]*匹配零个或多个首字母大写的字母\d匹配一个数字(?:\.\d+)*后面跟着点和数字，零次或多次-?匹配一个可选的结尾破折号\s+匹配一个或多个空白字符
英文:
You could use sub() as follows:
<!-- language: r -->
DF$indicator <- sub("^[A-Z]*\\d(?:\\.\\d+)*-?\\s+", "", DF$indicator)
DF
indicator amount
1 Total number customers per month (average) 12
2 Total of unique users served per month 45
3 Volume of merchandise sold per month 44
4 Quantity of bags received per month. 67
Here is an explanation of the regex pattern being used:
^from the start of the indicator[A-Z]*match zero or more leading capital letters\dmatch a digit(?:\.\d+)*followed by dot and digits, zero or more times-?match an optional ending dash\s+match one or more whitespace characters
答案2
得分: 1
Assuming that the indicator is always at the beginning and followed by a white space, you could do this to remove everything from the beginning to the first space.
library(stringr)
library(magrittr)
DF$indicator <- DF$indicator %>% str_remove_all("^.*? ")
> DF
indicator amount
1 每月客户总数（平均） 12
2 每月服务的唯一用户总数 45
3 每月销售的商品数量 44
4 每月收到的袋子数量 67
^锚定在字符串的开头.*匹配一切（直到空格）?在第一个空格处停止正则表达式匹配，而不是最后一个空格
英文:
Assuming that the indicator is always at the beginning and followed by a white space, you could do this to remove everything from the beginning to the first space.
library(stringr)
library(magrittr)
DF$indicator <- DF$indicator %>% str_remove_all("^.*? ")
> DF
indicator amount
1 Total number customers per month (average) 12
2 Total of unique users served per month 45
3 Volume of merchandise sold per month 44
4 Quantity of bags received per month. 67
^anchors to the beginning of the string.*matches everything (until the white space)?stops the regex at the first white space instead of the last white space
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