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dplyr's left_join keeps all of the rows on the left (a) and joins b, adding all the columns from b. This way, none of the information from the data frame that contains the information about the polygons is changed. That might solve this problem.

The command would be:

library(dplyr)
foo <- left_join(mydf, bar)

dplyr's left_join keeps all of the rows on the left (a) and joins b, adding all the columns from b. This way, none of the information from the data frame that contains the information about the polygons is changed. That might solve this problem.

The command would be:

library(dplyr)
foo <- left_join(mydf, bar)

Left join keeps all of the rows on the left (a) and joins b, adding all the columns from b. This way, none of the information from the data frame that contains the information about the polygons is changed. That might solve this problem.

The command would be:

library("dplyr")
foo <- left_join(mydf, bar)

dplyr's left_join keeps all of the rows on the left (a) and joins b, adding all the columns from b. This way, none of the information from the data frame that contains the information about the polygons is changed. That might solve this problem.

The command would be:

library(dplyr)
foo <- left_join(mydf, bar)

left_join(a,b) keeps all of the rows on the left (a) and joins b, adding all the columns from b. This way, none of the information from the data frame that contains the infomation about the polygons is changed.

Left join keeps all of the rows on the left (a) and joins b, adding all the columns from b. This way, none of the information from the data frame that contains the information about the polygons is changed. That might solve this problem.

The command would be:

library("dplyr")
foo <- left_join(mydf, bar)