将 tibble 中的每一行转换为列标题

Question

我正在尝试将第一列中的所有行转换为变量。稍后我可以使用 $ 表示法（例如 data$SWEAT_index）来调用它们。此时变量名称很长，稍后我将通过添加额外的列来简化它。也许我的方法是直截了当的。这个小问题怎么处理？

indices
# A tibble: 30 x 2
   Station                                 Value      
   <chr>                                   <chr>      
 1 Station identifier                      WMKC       
 2 Station number                          48615      
 3 Observation time                        190120/1200
 4 Station latitude                        6.16       
 5 Station longitude                       102.28     
 6 Station elevation                       5.0        
 7 Showalter index                         1.26       
 8 Lifted index                            -2.86      
 9 LIFT computed using virtual temperature -3.38      
10 SWEAT index                             187.99     
# ... with 20 more rows


data <- indices[-1,]
colnames(data) <-data[,1]
data
# A tibble: 29 x 2
   `c("Station number", "Observation time", "Station latitude", "Statio~ `c(48615, NA, 6.16, 102.28~
   <chr>                                                                                       <dbl>
 1 Station number                                                                           48615   
 2 Observation time                                                                            NA   
 3 Station latitude                                                                             6.16
 4 Station longitude                                                                          102.  
 5 Station elevation                                                                            5   
 6 Showalter index                                                                              1.26
 7 Lifted index                                                                                -2.86
 8 LIFT computed using virtual temperature                                                     -3.38
 9 SWEAT index                                                                                188.  
10 K index                                                                                     14.4 
# ... with 19 more rows

dput(indices)
structure(list(Station = c("Station identifier", "Station number", 
"Observation time", "Station latitude", "Station longitude", 
"Station elevation", "Showalter index", "Lifted index", "LIFT computed using virtual temperature", 
"SWEAT index", "K index", "Cross totals index", "Vertical totals index", 
"Totals totals index", "Convective Available Potential Energy", 
"CAPE using virtual temperature", "Convective Inhibition", "CINS using virtual temperature", 
"Equilibrum Level", "Equilibrum Level using virtual temperature", 
"Level of Free Convection", "LFCT using virtual temperature", 
"Bulk Richardson Number", "Bulk Richardson Number using CAPV", 
"Temp [K] of the Lifted Condensation Level", "Pres [hPa] of the Lifted Condensation Level", 
"Mean mixed layer potential temperature", "Mean mixed layer mixing ratio", 
"1000 hPa to 500 hPa thickness", "Precipitable water [mm] for entire sounding"
), Value = c(NA, 48615, NA, 6.16, 102.28, 5, 1.26, -2.86, -3.38, 
187.99, 14.4, 19, 23.9, 42.9, 409.13, 595.76, -26.9, -8.6, 228.72, 
226.79, 819.49, 871.25, 240, 349.48, 294.55, 938.33, 299.97, 
17.45, 5782, 46.56)), row.names = c(NA, -30L), class = c("tbl_df", 
"tbl", "data.frame"))

Answer 1

正如@NelsonGon 所说，我们可以使用spread

new_df <- tidyr::spread(indices, Station, Value)

现在您可以调用单个值，例如 new_df$`Station number、new_df$`Station identifier 等。

---

在基础 R 中，您可以转置，将其转​​换为数据框，然后使用 setNames 分配列名

new_df <- setNames(data.frame(t(indices$Value)), indices$Station)

但是，正如@Konrad Rudolph 提到的，转置数据帧可能会弄乱对象的数据类型，因此请小心处理。