合并两个条件，一个是列表

Question

I have two data frames that look something like this. Let's call the first data frame "Master"

row ID  Color
1   1   c("blue", "green")
2   1   red
3   2   red
4   3   c("pink", "blue", "purple")

Let's call the second data frame "Detail"

row ID  Color   Year
1   1   blue    2004
2   1   red     2000
3   1   green   2005
4   2   red     2005
5   3   pink    1999
6   3   brown   2008
7   3   blue    1997
8   3   pink    2007

I would like to add a column to Master that is the mean of the year values in Detail when two criteria are met:

1. ID matches (this is easy)
2. When Detail$Color is found in the list Master$Color. (this has proven to be difficult).

I have figured out that the command...

which(Detail$Color == Master$Color)

...will identify the Color pattern, but applying that command to either a merge or an apply statement has not worked out.

The result should look like this.

row ID  Color                       Mean_Year
1   1   c("blue", "green")          2004.5
2   1   red                         2000
3   2   red                         2005
4   3   c("pink", "blue", "purple") 2001

My real data has 10,000 rows in Master and 8,000,000 rows in Details, if that makes a difference.

Answer 1

I'm not sure what your data input file for the data frame 'Master' looks like, but let's assume that it contains one numeric vector (ID) and one character vector (Color) whereby each cell of the character vector contains one or more colours separated by a comma. If you import this into R, then you should get a data frame that looks like this:

row     ID     Color       
1       1      "blue, green"    
2       1      "red"    
3       2      "red"     
4       3      "pink, blue, purple"

I realise this is different from what's shown above for the data frame Master. Anyway, assuming that your input is similar to what I've just presented, once imported, the first thing to do is get rid of the white space in the vector master$Color

master$Color<-gsub(" ", "", master$Color)

Once that is done, it is relatively straightforward to do what you want:

ID.empty=NULL

for (i in 1:nrow(master)){ID.empty[i]=mean(detail$Year[detail$ID %in% master$ID[i] & is.element(detail$Color, unlist(c(strsplit(master$Color[i],','))))])}

print(ID.empty)

[1] 2004.5 2000.0 2005.0 2001.0