根据条件从长格式文件中只选择日期和相关行

Question

我有一个csv文件，可以导入到R. 这是一个数据帧，在“长格式”中有很多列，即有多个条目为相同的ID。我正在复制数据集的示例，以及我试图获得的结果数据集，仅使用前5列（实际数据中实际上有更多列）。从史蒂夫

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)) 

标识日期V1 V2 V3 V4 V5

1 ID1 30/10/20010：原始数据集在R中可以进行复制与此有关。 1 2 1 Y 0

2 id1 30/16/2005 from Anna。 09/08/2008来自Steve。 09/10/2009来自Steve。 NA NA NA N 0

3 id2 06/05/2004 from Allen。 1 2不适用不适用

4 id2 08/09/2005来自Anna。 1 NA 2 Y 0

5 id2 08/05/2008 from Allen。 30/10/2010从鲍比。 2 NA NA NA 0

6 id2 14/03/2002 from Steve。 23/03/2003从安娜。 NA NA 1 NA NA

7 id3 08/08/2002 from Steve。 1 2 1 Y 0

8 id3 08/08/2002 from Anna。 08/08/2002来自Steve。 2 NA NAŸNA

我想获得的数据集选择只与每个ID列date最近日期的行，如果这两个日期都一样，选择与行最少数目的NAS（的缺失值）：

标识日期V1 V2 V3 V4 V5

1 ID1 30/10/20010 1 2 1 Y 0

5 ID2 30/10/2010 2不适用不适用0

7 ID3 2002年8月8日1 2 1 Y 0

我想用 “从” 作为分隔符中的R拆分列，但是这并没有真正把我的任何地方：

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

后来我想，这可能是更好的选择BASH日期：但这也用不了我到任何地方：

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

基本上，我迷失在如何接近这个。 如果有人可以请建议正确的方法，希望只使用BASH或R，我将非常感激。 谢谢！

Answer 1

这里是您的示例数据工作

（后修正两个日期：30/10/20010 --> 30/10/2010和30/16/2005 --> 30/06/2005）的脚本

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) 

脚本的结果是这样的：

> 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