不带URI的SPARQL查询

如何解决不带URI的SPARQL查询

如果我写了tidyverse，那么我正在写一个SPARQL查询，但是如果我写了library(dplyr) # >= 1.0.0 library(tidyr) library(purrr) library(broom) lst1 <- list("AQ",c("AQ","WE","SZ"),"SZ","PO","LL")) nm1 <- grep("^E\\d+$",names(data),value = TRUE) fmlst <- do.call(c,lapply(lst1,function(vec) lapply(nm1,function(nm) reformulate(c(nm,vec),response = 'Y1')))) data %>% nest_by(SCHOOL,CLASS) %>% summarise(lmmodels = map(fmlst,~ lm(.x,data = data)),tidyout = map(lmmodels,tidy)) # A tibble: 36 x 4 # Groups: SCHOOL,CLASS [4] # SCHOOL CLASS lmmodels tidyout # <int> <int> <list> <list> # 1 1 2 <lm> <tibble [3 × 5]> # 2 1 2 <lm> <tibble [3 × 5]> # 3 1 2 <lm> <tibble [3 × 5]> # 4 1 2 <lm> <tibble [5 × 5]> # 5 1 2 <lm> <tibble [5 × 5]> # 6 1 2 <lm> <tibble [5 × 5]> # 7 1 2 <lm> <tibble [7 × 5]> # 8 1 2 <lm> <tibble [7 × 5]> # 9 1 2 <lm> <tibble [7 × 5]> #10 1 4 <lm> <tibble [3 × 5]> # … with 26 more rows ，那么它没有任何结果。该属性名称是motility，我已将数据上传到Jena Fuseki服务器中。

 FileInputStream inStream = new FileInputStream(new File(inputFilePath));
    Metadata Metadata = ImageMetadataReader.readMetadata(inStream);
    Bitmap picBitmap = BitmapFactory.decodeFile(inputFilePath);
    FileOutputStream outStream = new FileOutputStream(new File(copiedFile));
    picBitmap.compress(Bitmap.CompressFormat.JPEG,100,outStream);
    for (Directory directory : Metadata.getDirectories()) {
        for (Tag tag : directory.getTags()) {
            System.out.println("Tag :" + tag);
        }
    }

其背后的逻辑或语法错误是什么？