0

我有一个 csv 文件，如下所示：

> collar<-read.csv("41361_predicted_by_41365_cluster.csv",header=TRUE,stringsAsFactors = F)
> head(collar)
        observed      predicted probability
1 Moving/Feeding       Standing           1
2       Standing Feeding/Moving           1
3       Standing Feeding/Moving           1
4       Standing Feeding/Moving           1
5       Standing Feeding/Moving           1
6       Standing Feeding/Moving           1

我的问题很简单，但实施起来有点困难，因为我正在学习 R。我只需要在列中进行以下字符替换predicted：

1) 替换Feeding/Moving为Standing

2）替换Moving/Feeding为Feeding/Moving

3）替换Standing为Moving/Feeding

然后我想使用名称以 csv 格式写入数据"corrected_41361_predicted_by_41365_cluster"。

我正在寻找最紧凑的方式。有没有不同的方法gsub()？任何输入表示赞赏！

4

2 回答 2

2

像那样？（我稍微改变了你的样本，包括“移动/喂食”）

library(dplyr)

df %>%
  mutate(predicted = case_when(predicted == 'Feeding/Moving' ~ 'Standing',
                               predicted == 'Moving/Feeding' ~ 'Feeding/Moving',
                               predicted == 'Standing' ~ 'Moving/Feeding',
                               TRUE ~ predicted)) %>%
 write.csv(file = "corrected_41361_predicted_by_41365_cluster.csv")

该文件包含结果：


        observed      predicted probability
1 Moving/Feeding Moving/Feeding           1
2       Standing       Standing           1
3       Standing       Standing           1
4       Standing       Standing           1
5       Standing Feeding/Moving           1
6       Standing       Standing           1
于 2019-04-02T10:35:09.790 回答
1

你可以尝试这样的事情：

# Read all files in folder
mydocpath = "C:/Users/yourdata"
files <- list.files(path=sprintf("%s",mydocpath), pattern="*.csv", full.names=TRUE, recursive=FALSE)
print(files)

# Read to "list of DFs"
myf = lapply(files, function(x) data.frame(read.csv(x, sep=";", header=F, na = "na", skip=1)))
# Access single DF in list
myf[[1]]

# Do something to each DF
for(i in myf){
  print(head(myf))
}
于 2019-04-02T10:20:55.420 回答