Question

我有一个如下的数据帧，它是分类器的输出。

col1, class
 123, 2
 122, 5
 124, 7
 125, 9
 126, 15
 127, 2
 128, 19
 129, 5
 130, 7
 179, 9
 180, 3

我想找到具有某种类型模式的行，就像其类在seq 5,7,9中的所有行一样。

我提出的解决方案是通过移动一行并比较列来粘贴类列

 col1, class, class1, class2
 123, 2,5,7
 122, 5,7,9
 124, 7,9,15
 125, 9,15,2
 126, 15,2,19
 127, 2,19,5
 128, 19,5,7
 129, 5,7,9
 130, 7,9,3
 179, 9,3,NA,
 180, 3,NA,NA

只有当我的模式中的字段数相同时，才能解决这个问题。有些模式甚至可以有5到7个字段。

Answer 1

我们可以使用shift中的data.table，然后将paste元素放在一起，并查看我们的位置579

n <- 3
library(data.table)
setDT(df1)[, which(do.call(paste0, shift(class, seq(n)-1, type = "lead"))=="579")]
#[1] 2 8

或者paste代替Map我们可以将Reduce与setDT(df1)[, which(Reduce(`&`, Map(`==`, shift(class, seq(n)-1, type = "lead"), c(5, 7, 9))))] #[1] 2 8

一起使用

public class MainActivity extends ListActivity {
Cursor cursor;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    String[] columns = { MediaStore.Audio.Media._ID };
    int[] displayViews = new int[] {android.R.id.text1};
    ContentResolver cr = this.getContentResolver();
    cursor = cr.query(MediaStore.Audio.Albums.EXTERNAL_CONTENT_URI,columns, null, null, null);
    setListAdapter(new SimpleCursorAdapter(this,android.R.layout.simple_list_item_1,cursor,columns,displayViews,1));
}

public static ArrayList<SongDto> getMusicInfos(Context context) {

    ArrayList<SongDto> musicInfos = new ArrayList<SongDto>();

    Cursor cursor = context.getContentResolver().query(
            MediaStore.Audio.Media.EXTERNAL_CONTENT_URI, null, null, null,
            MediaStore.Audio.Media.DEFAULT_SORT_ORDER);
    if (cursor == null) {
        return null;
    }


    for (int i = 0; i < cursor.getCount(); i++) {
        cursor.moveToNext();


        int isMusic = cursor.getInt(cursor
                .getColumnIndex(MediaStore.Audio.Media.IS_MUSIC));

        if (isMusic != 0) {
            SongDto music = new SongDto();


            music.path = cursor.getString(cursor
                    .getColumnIndexOrThrow(MediaStore.Audio.Media.DATA));

            if (!new File(music.path).exists()) {
                continue;
            }


            music.songId = cursor.getLong(cursor
                    .getColumnIndexOrThrow(MediaStore.Audio.Media._ID));

            music.songTitle = cursor.getString(cursor
                    .getColumnIndexOrThrow(MediaStore.Audio.Media.TITLE));


            music.songTitle = cursor.getString(cursor
                    .getColumnIndex(MediaStore.Audio.Media.DISPLAY_NAME));


            music.album = cursor.getString(cursor
                    .getColumnIndexOrThrow(MediaStore.Audio.Media.ALBUM));


            music.songArtist = cursor.getString(cursor
                    .getColumnIndexOrThrow(MediaStore.Audio.Media.ARTIST));

            music.duration = cursor
                    .getLong(cursor
                            .getColumnIndexOrThrow(MediaStore.Audio.Media.DURATION));

            MediaMetadataRetriever mmr = new MediaMetadataRetriever();
            mmr.setDataSource(music.path);
            mmr.release();

            musicInfos.add(music);
        }
    }

    return musicInfos;
}
}

Answer 2

一个更长的基础R替代方案，原则上类似于@ akrun的答案：

which(do.call(paste0, cbind(df1, with(df1, class[seq_along(class)+1]),
                             with(df1, class[seq_along(class)+2]))[-1]) == "579")
#[1] 2 8

数据：

df1 <- structure(list(col1 = c(123L, 122L, 124L, 125L, 126L, 127L, 128L, 129L, 130L, 179L, 180L), class = c(2L, 5L, 7L, 9L, 15L, 2L, 19L, 5L, 7L, 9L, 3L)), .Names = c("col1", "class"), class = "data.frame", row.names = c(NA, -11L))

查找R数据帧中的行，其中列值遵循序列

2 个答案: