如何解决如何在给定年份用NA替换全0的列？
以下数据集有一些年份，其中的列全部或部分填充零。我试图用NA替换给定年份中全为零的列。包括我在下面尝试过的代码。
year tract Black MedianRent Moved in 2015 or later Moved in 2017 or later Built 2010 or later SNAP benefits Bachelor's degree
1 2018 Census Tract 401 21 919 0 0 0 91831 2071
2 2018 Census Tract 402 0 882 0 0 0 90117 1217
3 2018 Census Tract 403 37 749 0 0 0 94611 1479
4 2018 Census Tract 404 85 908 0 0 0 97322 2272
5 2018 Census Tract 405 5 810 0 0 0 76444 1731
6 2018 Census Tract 406 0 946 0 0 0 122761 581
years <- unique(final$year)
final %>%
group_by(year) %>%
mutate_at(final[,c(2)],~ if (all(. == 0)) . else NA_real_) %>%
ungroup()
years <- unique(final$year)
bd1 <- data.frame()
bd2 <- data.frame()
for (i in years){
bd1 <- final %>%
filter(year == i)
bd1 <- bd1 %>%
na_if(all(0))
bd2 <- smartbind(bd1,bd2)
}
final2 <- final %>%
group_by(year) %>%
mutate_at(vars(),~ if (any(. != 0)) . else NA_real_) %>%
ungroup()
解决方法
尝试以下方法：
library(dplyr)
final %>%
group_by(year) %>%
mutate(across(-2,~if (all(. == 0)) NA_real_ else .)) %>%
#With mutate_at
#mutate_at(-2,~if (all(. == 0)) NA_real_ else .) %>%
ungroup()
如果year的值为{0，则此值将给定NA的列值替换为all。
这项工作：
> library(dplyr)
> rent %>% group_by(year) %>% mutate(across(where(is.numeric),~ replace(.,sum(.) == 0,NA)))
# A tibble: 6 x 9
# Groups: year [1]
year tract Black MedianRent `Moved in 2015 or l~ `Moved in 2017 or l~ `Built 2010 or la~ `SNAP benefits` `Bachelor's degr~
<dbl> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 2018 Census Tra~ 21 919 NA NA NA 91831 2071
2 2018 Census Tra~ 0 882 NA NA NA 90117 1217
3 2018 Census Tra~ 37 749 NA NA NA 94611 1479
4 2018 Census Tra~ 85 908 NA NA NA 97322 2272
5 2018 Census Tra~ 5 810 NA NA NA 76444 1731
6 2018 Census Tra~ 0 946 NA NA NA 122761 581
>
使用的数据：
structure(list(year = c(2018,2018,2018),tract = c("Census Tract 401","Census Tract 402","Census Tract 403","Census Tract 404","Census Tract 405","Census Tract 406"
),Black = c(21,37,85,5,0),MedianRent = c(919,882,749,908,810,946),`Moved in 2015 or later` = c(0,`Moved in 2017 or later` = c(0,`Built 2010 or later` = c(0,`SNAP benefits` = c(91831,90117,94611,97322,76444,122761),`Bachelor's degree` = c(2071,1217,1479,2272,1731,581)),row.names = c(NA,-6L),class = c("tbl_df","tbl","data.frame"))
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