Question

我为一些调查问题生成一些频率，然后将其中一些问题放到一个数据框中。每个问题的回答为是/否，也报告为No %和Yes %。

现在，如果在给定的行No < 15或Yes < 15中，则该行中只有合计值可见，而“否”，“是”，No %和Yes %列中被屏蔽为NA。

我正在case_when和其他选项上乱七八糟，但是运气不好。我会插手，但如果有一个明显的解决方案能使人满意，我将不胜感激。我不愿意dplyr寻求解决方案。预先感谢！

示例数据帧在下面显示为mytab：

mytab <- structure(list(No = c(271L, 1395L, 1393L, 1338L, 1254L, 1355L, 1332L, 1380L, 1360L), Yes = c(1138L, 14L, 16L, 71L, 155L, 54L, 77L, 29L, 49L),
               Total = c(1409, 1409, 1409, 1409, 1409, 1409, 1409, 1409, 1409),
               `No (%)` = c(19.2334989354152, 99.0063875088715, 98.8644428672818, 94.9609652235628, 88.9992902767921, 96.1674946770759, 94.5351312987935, 97.9418026969482, 96.5223562810504),
               `Yes (%)` = c(80.7665010645848, 0.99361249112846, 1.13555713271824, 5.03903477643719, 11.0007097232079, 3.83250532292406, 5.46486870120653, 2.05819730305181, 3.47764371894961)),
          row.names = c(NA, -9L),
          class = "data.frame")

mytab

#>     No  Yes Total   No (%)    Yes (%)
#> 1  271 1138  1409 19.23350 80.7665011
#> 2 1395   14  1409 99.00639  0.9936125
#> 3 1393   16  1409 98.86444  1.1355571
#> 4 1338   71  1409 94.96097  5.0390348
#> 5 1254  155  1409 88.99929 11.0007097
#> 6 1355   54  1409 96.16749  3.8325053
#> 7 1332   77  1409 94.53513  5.4648687
#> 8 1380   29  1409 97.94180  2.0581973
#> 9 1360   49  1409 96.52236  3.4776437

解决方案应产生mytab2，然后可以将其通过管道传输到knitr。

mytab2 <- structure(list(No = c(271L, NA, 1393L, 1338L, 1254L, 1355L, 1332L, 1380L, 1360L),
                         Yes = c(1138L, NA, 16L, 71L, 155L, 54L, 77L, 29L, 49L),
                         Total = c(1409, 1409, 1409, 1409, 1409, 1409, 1409, 1409, 1409),
                         `No (%)` = c(19.2334989354152, NA, 98.8644428672818, 94.9609652235628, 88.9992902767921, 96.1674946770759, 94.5351312987935, 97.9418026969482, 96.5223562810504),
                         `Yes (%)` = c(80.7665010645848, NA, 1.13555713271824, 5.03903477643719, 11.0007097232079, 3.83250532292406, 5.46486870120653, 2.05819730305181, 3.47764371894961)),
                    row.names = c(NA, -9L),
                    class = "data.frame")

mytab2
#>     No  Yes Total   No (%)   Yes (%)
#> 1  271 1138  1409 19.23350 80.766501
#> 2   NA   NA  1409       NA        NA
#> 3 1393   16  1409 98.86444  1.135557
#> 4 1338   71  1409 94.96097  5.039035
#> 5 1254  155  1409 88.99929 11.000710
#> 6 1355   54  1409 96.16749  3.832505
#> 7 1332   77  1409 94.53513  5.464869
#> 8 1380   29  1409 97.94180  2.058197
#> 9 1360   49  1409 96.52236  3.477644

Answer 1

这与divibisan的答案相同，但具有data.table语法，该语法减少了表名的重复并使用between（因为它似乎合适）：</ p>

library(data.table)
mybadtab = data.table(mytab)

mymin = 15
badcols = c("No", "Yes", "No (%)", "Yes (%)")
mybadtab[!( No %between% c(mymin, Total - mymin) ), (badcols) := NA]

     No  Yes Total   No (%)   Yes (%)
1:  271 1138  1409 19.23350 80.766501
2:   NA   NA  1409       NA        NA
3: 1393   16  1409 98.86444  1.135557
4: 1338   71  1409 94.96097  5.039035
5: 1254  155  1409 88.99929 11.000710
6: 1355   54  1409 96.16749  3.832505
7: 1332   77  1409 94.53513  5.464869
8: 1380   29  1409 97.94180  2.058197
9: 1360   49  1409 96.52236  3.477644

管道形式...

library(magrittr)
library(knitr)

mymin = 15
badcols = c("No", "Yes", "No (%)", "Yes (%)")

data.table(mytab)[!( No %between% c(mymin, Total - mymin) ), (badcols) := NA] %>% 
  kable

|   No|  Yes| Total|   No (%)|   Yes (%)|
|----:|----:|-----:|--------:|---------:|
|  271| 1138|  1409| 19.23350| 80.766501|
|   NA|   NA|  1409|       NA|        NA|
| 1393|   16|  1409| 98.86444|  1.135557|
| 1338|   71|  1409| 94.96097|  5.039035|
| 1254|  155|  1409| 88.99929| 11.000710|
| 1355|   54|  1409| 96.16749|  3.832505|
| 1332|   77|  1409| 94.53513|  5.464869|
| 1380|   29|  1409| 97.94180|  2.058197|
| 1360|   49|  1409| 96.52236|  3.477644|

Answer 2

在基数R中，您可以仅使用带有方括号的子集来获取适当的行，然后将NA分配给要更改的列。注意：这将修改 mytab的值。如果要在新的data.frame中进行更改，则需要复制mytab并修改副本：

mytab2 <- mytab
mytab2[mytab2$No < 15 | mytab2$Yes < 15, c('No', 'Yes', 'No (%)', 'Yes (%)')] <- NA
mytab2
    No  Yes Total   No (%)   Yes (%)
1  271 1138  1409 19.23350 80.766501
2   NA   NA  1409       NA        NA
3 1393   16  1409 98.86444  1.135557
4 1338   71  1409 94.96097  5.039035
5 1254  155  1409 88.99929 11.000710
6 1355   54  1409 96.16749  3.832505
7 1332   77  1409 94.53513  5.464869
8 1380   29  1409 97.94180  2.058197
9 1360   49  1409 96.52236  3.477644

Answer 3

尝试一下：

df<-as.data.frame(list(No = c(271, 1395, 1393, 1338, 1254, 1355, 1332, 1380, 1360),
                   Yes = c(1138, 14, 16, 71, 155, 54, 77, 29, 49),
                   Total = c(1409, 1409, 1409, 1409, 1409, 1409, 1409, 1409, 1409)))

df$NoPct<-0
df$YesPct<-0

rowcalc<-function(x){
  if (x[1]<15 | x[2]<15){
      x[1]= x[2]= x[4]=x[5]=NA
  } else {
         x[4]<- round(100*x[1]/x[3],digits=2) #rounding to 2 decimal places 
         x[5]<- round(100*x[2]/x[3],digits=2) 
          }
        return(x)
        }
t(apply(df,1,rowcalc)) #apply rowcalc to every row & transpose it  

#      No  Yes Total NoPct YesPct
#[1,]  271 1138  1409 19.23  80.77
#[2,]   NA   NA  1409    NA     NA
#[3,] 1393   16  1409 98.86   1.14
#[4,] 1338   71  1409 94.96   5.04
#[5,] 1254  155  1409 89.00  11.00
#[6,] 1355   54  1409 96.17   3.83
#[7,] 1332   77  1409 94.54   5.46
#[8,] 1380   29  1409 97.94   2.06
#[9,] 1360   49  1409 96.52   3.48

屏蔽满足条件的行中的特定单元格值

3 个答案: