我正在尝试内部连接两个数据集：50000 obs的df1看起来像这样：
Name | Line.1 | Line.2 | Town | County | Postcode
-------------------|------------------|------------|------------|--------------|----------
ACME Inc | 63 Long Street | | Fakeington | Lincolnshire | PA4 8QU
BETA LTD | 91a | Main Drove | Cloud City | Something | BN1 6LD
The Giga | 344 Lorem Street | | Ipsom | Dolor | G2 8LY
500000 obs的df2如下所示：
Name | AddressLine1 | AddressLine2 | AddressLine3 | AddressLine4 | Postcode | RatingValue
-------------------|----------------|------------------|--------------|--------------|----------|-------------
ACME | | 63 Long Street | Fakeington | Lincolnshire | PA4 8QU | 1
Random Company | | Rose Ave | Fakeington | | AB2 51GL | 5
BETA Limited | Business House | 91a Main Drove | Something | | BN1 6LD | 3
Giga Incorporated | | 344 Lorem Street | Ipsum | Dolor | G2 8LY | 5
我想去df_final。
Name | Postcode | RatingValue
-------------------|----------|-------------
ACME Inc | PA4 8QU | 1
BETA LTD | BN1 6LD | 3
Giga Incorporated | G2 8LY | 5
这些是一对一的匹配，并且df1中的所有值都应该存在于df2中。Postcode是完全匹配的，而地址被分割成多行，没有规则模式，所以我认为我最好的选择是通过Name匹配。
我尝试了fuzzyjoin包，但我得到了Error: cannot allocate vector of size 120.6 Gb，所以我想我必须使用另一种适用于更大数据集的方法。
关于解决这个问题的最佳方法是什么，有什么想法吗？
df1 <- data.frame(
stringsAsFactors = FALSE,
Name = c("ACME Inc", "BETA LTD", "Giga Incorporated"),
Line.1 = c("63 Long Street", "91a", "344 Lorem Street"),
Line.2 = c(NA, "Main Drove", NA),
Town = c("Fakeington", "Cloud City", "Ipsom"),
County = c("Lincolnshire", "Something", "Dolor"),
Postcode = c("PA4 8QU", "BN1 6LD", "G2 8LY")
)
df2 <- data.frame(
stringsAsFactors = FALSE,
Name = c("ACME", "Random Company","BETA Limited","Giga Incorporated"),
AddressLine1 = c(NA, NA, "Business House", NA),
AddressLine2 = c("63 Long Street", "Rose Ave","91a Main Drove","344 Lorem Street"),
AddressLine3 = c("Fakeington", "Fakeington", "Something", "Ipsum"),
AddressLine4 = c("Lincolnshire", NA, NA, "Dolor"),
Postcode = c("PA4 8QU", "AB2 51GL", "BN1 6LD", "G2 8LY"),
RatingValue = c(1L, 5L, 3L, 5L)
)
也许下面的内容可以满足问题的要求。它使用包stringdist，而不是fuzzyjoin。
首先，由于匹配是精确的，因此仅使用merge和Postcode。然后得出Name的相似性。如果它们高于预定阈值，则保留这些行。
thresh <- 0.75
df_final <- merge(df2[c(1, 6:7)], df1[c(1, 6)], by = "Postcode", suffixes = c("",".y"))
i <- apply(df_final[c(2, 4)], 1, function(x) {stringdist::stringsim(x[1], x[2], method = 'jw')}) >= thresh
df_final <- df_final[i, c(2, 1, 3)]
df_final
# Name Postcode RatingValue
#1 BETA Limited BN1 6LD 3
#2 Giga Incorporated G2 8LY 5
#3 ACME PA4 8QU 1