我有两个数据集，如下所示：
DT1 <- structure(list(Province = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2,
2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3), Year = c(2000,
2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000,
2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000, 2001, 2001,
2001, 2002, 2002, 2002), Municipality = c("Something", "Anything",
"Nothing", "Something", "Anything", "Nothing", "Something", "Anything",
"Nothing", "Something", "Anything", "Nothing", "Something", "Anything",
"Nothing", "Something", "Anything", "Nothing", "Something", "Anything",
"Nothing", "Something", "Anything", "Nothing", "Something", "Anything",
"Nothing"), Values = c(0.59, 0.58, 0.66, 0.53, 0.94, 0.2, 0.86,
0.85, 0.99, 0.59, 0.58, 0.66, 0.53, 0.94, 0.2, 0.86, 0.85, 0.99,
0.59, 0.58, 0.66, 0.53, 0.94, 0.2, 0.86, 0.85, 0.99)), row.names = c(NA,
-27L), class = c("tbl_df", "tbl", "data.frame"))
DT2 <- structure(list(Province = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2,
2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3), Year = c(2000,
2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000,
2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000, 2001, 2001,
2001, 2002, 2002, 2002), Municipality = c("Some", "Anything",
"Nothing", "Someth.", "Anything", "Not", "Something", "Anything",
"None", "Some", "Anything", "Nothing", "Someth.", "Anything",
"Not", "Something", "Anything", "None", "Some", "Anything", "Nothing",
"Someth.", "Anything", "Not", "Something", "Anything", "None"
), `Other Values` = c(0.41, 0.42, 0.34, 0.47, 0.0600000000000001,
0.8, 0.14, 0.15, 0.01, 0.41, 0.42, 0.34, 0.47, 0.0600000000000001,
0.8, 0.14, 0.15, 0.01, 0.41, 0.42, 0.34, 0.47, 0.0600000000000001,
0.8, 0.14, 0.15, 0.01)), row.names = c(NA, -27L), class = c("tbl_df",
"tbl", "data.frame"))
我试图将它们匹配如下，这是叶在这个链接中建议的。
library(fuzzyjoin); library(dplyr);
stringdist_join(DT1, DT2,
by = "Municipality",
mode = "left",
ignore_case = TRUE,
method = "jw",
max_dist = 10,
distance_col = "dist") %>%
group_by(Municipality.x) %>%
top_n(1, -dist)
问题是，这些代码完全毁了我的电脑，所以我想把代码分成几组，以限制字符串比较的数量。我试过了：
library(fuzzyjoin); library(dplyr);
stringdist_join(DT1, DT2,
by = c("Municipality","Year", "State"),
mode = "left",
ignore_case = TRUE,
method = "jw",
max_dist = 10,
distance_col = "dist") %>%
group_by(Municipality.x) %>%
top_n(1, -dist)
stringdist_join(DT1, DT2,
by = "Municipality",
mode = "left",
ignore_case = TRUE,
method = "jw",
max_dist = 10,
distance_col = "dist") %>%
group_by(Municipality, Year, Province) %>%
top_n(1, -dist)
但两者都给了我以下各自的错误：
Error: All columns in a tibble must be vectors.
x Column `col` is NULL.
Run `rlang::last_error()` to see where the error occurred.
和：
Error: Must group by variables found in `.data`.
* Column `Municipality` is not found.
* Column `Year` is not found.
* Column `Province` is not found.
Run `rlang::last_error()` to see where the error occurred.
这样做的正确方法是什么？
您走在了正确的轨道上——只有几个打字错误/错误，您需要完成列名的更改/替换。
此外，在你的第一次面试中，你需要弄清楚你想如何选择"；最佳匹配"；基于Municipality.dist、Province.dist和Year.dist.
如果你先把年份和省份弄清楚，也许第二种方法效果更好。
DT1 <- structure(list(Province = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3), Year = c(2000, 2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002), Municipality = c("Something", "Anything", "Nothing", "Something", "Anything", "Nothing", "Something", "Anything", "Nothing", "Something", "Anything", "Nothing", "Something", "Anything", "Nothing", "Something", "Anything", "Nothing", "Something", "Anything", "Nothing", "Something", "Anything", "Nothing", "Something", "Anything", "Nothing"), Values = c(0.59, 0.58, 0.66, 0.53, 0.94, 0.2, 0.86, 0.85, 0.99, 0.59, 0.58, 0.66, 0.53, 0.94, 0.2, 0.86, 0.85, 0.99, 0.59, 0.58, 0.66, 0.53, 0.94, 0.2, 0.86, 0.85, 0.99)), row.names = c(NA, -27L), class = c("tbl_df", "tbl", "data.frame"))
DT2 <- structure(list(Province = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3), Year = c(2000, 2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002), Municipality = c("Some", "Anything", "Nothing", "Someth.", "Anything", "Not", "Something", "Anything", "None", "Some", "Anything", "Nothing", "Someth.", "Anything", "Not", "Something", "Anything", "None", "Some", "Anything", "Nothing", "Someth.", "Anything", "Not", "Something", "Anything", "None"), `Other Values` = c(0.41, 0.42, 0.34, 0.47, 0.0600000000000001, 0.8, 0.14, 0.15, 0.01, 0.41, 0.42, 0.34, 0.47, 0.0600000000000001, 0.8, 0.14, 0.15, 0.01, 0.41, 0.42, 0.34, 0.47, 0.0600000000000001, 0.8, 0.14, 0.15, 0.01)), row.names = c(NA, -27L), class = c("tbl_df", "tbl", "data.frame"))
library(fuzzyjoin); library(dplyr);
stringdist_join(DT1, DT2,
by = c("Municipality", "Year", "Province"),
mode = "left",
ignore_case = TRUE,
method = "jw",
max_dist = 10,
distance_col = "dist") %>%
group_by(Municipality.x) %>%
slice_min(Municipality.dist)
#> # A tibble: 135 x 12
#> # Groups: Municipality.x [3]
#> Province.x Year.x Municipality.x Values Province.y Year.y Municipality.y
#> <dbl> <dbl> <chr> <dbl> <dbl> <dbl> <chr>
#> 1 1 2000 Anything 0.580 1 2000 Anything
#> 2 1 2000 Anything 0.580 1 2001 Anything
#> 3 1 2000 Anything 0.580 1 2002 Anything
#> 4 1 2000 Anything 0.580 2 2000 Anything
#> 5 1 2000 Anything 0.580 2 2001 Anything
#> 6 1 2000 Anything 0.580 2 2002 Anything
#> 7 1 2000 Anything 0.580 3 2000 Anything
#> 8 1 2000 Anything 0.580 3 2001 Anything
#> 9 1 2000 Anything 0.580 3 2002 Anything
#> 10 1 2001 Anything 0.94 1 2000 Anything
#> # ... with 125 more rows, and 5 more variables: `Other Values` <dbl>,
#> # Municipality.dist <dbl>, Province.dist <dbl>, Year.dist <dbl>, dist <lgl>
stringdist_join(DT1, DT2,
by = "Municipality",
mode = "left",
ignore_case = TRUE,
method = "jw",
max_dist = 10,
distance_col = "dist") %>%
group_by(Municipality.x, Year.x, Province.x) %>%
slice_min(dist)
#> # A tibble: 135 x 9
#> # Groups: Municipality.x, Year.x, Province.x [27]
#> Province.x Year.x Municipality.x Values Province.y Year.y Municipality.y
#> <dbl> <dbl> <chr> <dbl> <dbl> <dbl> <chr>
#> 1 1 2000 Anything 0.580 1 2000 Anything
#> 2 1 2000 Anything 0.580 1 2001 Anything
#> 3 1 2000 Anything 0.580 1 2002 Anything
#> 4 1 2000 Anything 0.580 2 2000 Anything
#> 5 1 2000 Anything 0.580 2 2001 Anything
#> 6 1 2000 Anything 0.580 2 2002 Anything
#> 7 1 2000 Anything 0.580 3 2000 Anything
#> 8 1 2000 Anything 0.580 3 2001 Anything
#> 9 1 2000 Anything 0.580 3 2002 Anything
#> 10 2 2000 Anything 0.580 1 2000 Anything
#> # ... with 125 more rows, and 2 more variables: `Other Values` <dbl>,
#> # dist <dbl>
创建于2020-12-07由reprex包(v0.3.0(