R 根据长格式文件中的条件，仅选择日期和相关行

我有一个csv文件，可以导入到R。 这是一个“长格式”的数据帧，有许多列，即同一ID有多个条目。我正在复制一个数据集示例和我试图获得的结果数据集，仅使用前5列（我的真实数据中实际上有更多的列）。原始数据集可通过以下方式在R中复制：

df <- data.frame(id = c("id1","id1","id2","id2","id2","id2","id3","id3"), date = c("30/10/20010 from Steve.","30/16/2005 from Anna. 09/08/2008 from Steve. 09/10/2009 from Steve.","06/05/2004 from Allen.","08/09/2005 from Anna.","08/05/2008 from Allen. 30/10/2010 from Bobby.","14/03/2002 from Steve. 23/07/2003 from Anna.","08/08/2002 from Steve.", "08/08/2002 from Anna. 08/08/2002 from Steve."), v1 = c(1,NA,1,1,2,NA,1,2), v2 = c(2,NA,2,NA,NA,NA,2,NA), v3 = c(1,NA,NA,2,NA,1,1,NA), v4 = c("Y","N","N","Y","NA","NA","Y","Y"), v5 = c(0,0,NA,0,0,NA,0,NA))

然后我想在BASH中选择日期可能更好：但这也不能带我去任何地方：

try<- strsplit(df$date, "from", fixed=TRUE)

grep "[0-9][0-9]\/[0-9][0-9]\/20[0-9][0-9]" file.csv | less -S

基本上，我不知道如何处理这个问题。 如果有人能建议正确的方法，我将非常感激，希望只是使用BASH或R？

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谢谢大家!

下面是一个用于示例数据的脚本

（更正两个日期后：

30/10/20010-->30/10/2010

和

30/16/2005-->30/06/2005

）

---

下面是一个用于示例数据的脚本

（更正两个日期后：

30/10/20010-->30/10/2010

和

30/16/2005-->30/06/2005

）

---

亲爱的yellowcap，非常感谢您的帮助，代码有效，此步骤后只有一些警告：“df.dates再次您好，我刚刚重新检查了这个，它给了我很多错误，问题是当我没有唯一的日期以及相同ID中的重复日期时，因此，例如，如果我添加第四个ID，如下所示：df亲爱的yellowcap，非常感谢您的帮助，代码有效，此步骤后只有一些警告：“df.dates您好，我刚刚重新检查了它，它给了我很多错误，问题是当我没有唯一的日期以及相同ID中的重复日期时，例如，如果我添加第四个ID，如：df

df <- data.frame(id = c("id1","id1","id2","id2","id2","id2","id3","id3"), date = c("30/10/2010 from Steve.","30/06/2005 from Anna. 09/08/2008 from Steve. 09/10/2009 from Steve.","06/05/2004 from Allen.","08/09/2005 from Anna.","08/05/2008 from Allen. 30/10/2010 from Bobby.","14/03/2002 from Steve. 23/07/2003 from Anna.","08/08/2002 from Steve.", "08/08/2002 from Anna. 08/08/2002 from Steve."), v1 = c(1,NA,1,1,2,NA,1,2), v2 = c(2,NA,2,NA,NA,NA,2,NA), v3 = c(1,NA,NA,2,NA,1,1,NA), v4 = c("Y","N","N","Y","NA","NA","Y","Y"), v5 = c(0,0,NA,0,0,NA,0,NA))

# Convert string to list of dates, extract maximum (earliest date)
df.dates <- sapply(as.character(df$date), function(x) strsplit(x, "\\."))
df.dates <- lapply(df.dates, function(x) as.Date(x, format='%d/%m/%Y'))
df.dates <- lapply(df.dates, max)

# Add to dataframe
df$latest <- unlist(df.dates)

# Count the number of NA values per row
df$naCount <- apply(df, 1, function(x) sum(is.na(x[3:7])))

# Split data and select either maximum (latest) date or minimal NA count
df.split <- split(df, df$id)

df.select <- lapply(df.split, function(x){

    # Select by given criterion
    if(length(unique(x$latest)) == 1){
        y <- x[which(min(x$naCount) == x$naCount),]
        }
    else{
        y <- x[which(max(x$latest) == x$latest),]
    }

    # Check if selection was successful
    if(nrow(y) != 1) cat('Warning: non-unique choice, returning more than one line')

    # Return result
    y
})

# Combine into output dataframe
df.select <- do.call(rbind, df.select)

> df.select

     id                                          date v1 v2 v3 v4 v5 latest naCount
id1 id1                        30/10/2010 from Steve.  1  2  1  Y  0  14912       0
id2 id2 08/05/2008 from Allen. 30/10/2010 from Bobby.  2 NA NA NA  0  14912       2
id3 id3                        08/08/2002 from Steve.  1  2  1  Y  0  11907       0