我有一个看似简单的问题,但我无法弄清楚如何得到我想要的东西.
我的数据如下:
Job C/C++ Java Python
Student FALSE TRUE FALSE
Developer TRUE TRUE TRUE
Developer TRUE TRUE FALSE
Sysadmin TRUE FALSE FALSE
Student FALSE TRUE TRUE
我想按"作业"列进行分组,并计算TRUE每列中的s 数.我想要的输出看起来像这样:
Job C/C++ Java Python
Student 0 2 1
Developer 2 2 1
Sysadmin 1 0 0
任何帮助将不胜感激.
请原谅我的新手问题.我试过寻找答案,但鉴于关键字,搜索这类事情是相当困难的......
我使用Lucene 5.2.x索引一组文档,每个文档有两个字段:id和description.
我ids从系统中的先前查询中获取了一组.现在,我想获得Lucene文本搜索结果,description但只能从集合中的文档中获取ids.如果我在MySQL中这样做(天真地),我可能会这样做:
SELECT * FROM mytable
WHERE description LIKE 'blah%'
AND
id IN (6345, 5759, 333, ...)
这组ids可能数以万计.Lucene最好的解决方法是什么？我可以构建一个Lucene查询来有效地处理这个问题,还是应该搜索整个文档索引然后进行集合交集？别的什么？
谢谢!
假设我有以下数据框:
Base Coupled Derived Decl
1 0 0 1
1 7 0 1
1 1 0 1
2 3 12 1
1 0 4 1
这是dput输出:
temp <- structure(list(Base = c(1L, 1L, 1L, 2L, 1L), Coupled = c(0L,7L, 1L, 3L, 0L), Derived = c(0L, 0L, 0L, 12L, 4L), Decl = c(1L, 1L, 1L, 1L, 1L)), .Names = c("Base", "Coupled", "Derived", "Decl"), row.names = c(NA, 5L), class = "data.frame")
我想计算每列的中位数.然后,对于每一行,我想计算大于其各自列的中位数的单元格值的数量,并将其作为名为AboveMedians的列附加.
在这个例子中,中位数就是c(1,1,0,1).我想要的结果表是
Base Coupled Derived Decl AboveMedians
1 …R新手在这里,所以请原谅我的无知.我的数据如下:
JOB_ROLE EXP_IT_NETW
1 Software engineering-related (developer, tester, project manager, architecture) 5<10
3 See below None
4 Student <1
5 Software engineering-related (developer, tester, project manager, architecture) 1<5
6 Blogger 10+
我想计算第2列中每个值的实例,并按第1列对计数进行分组,以便结果如下所示:
JOB_ROLE None <1 1<5 5<10 10+
Software engineer 3 5 10 15 3
Student 10 7 5 1 0
...
关于如何做到这一点的任何想法？我的输出输出低于.提前致谢!
structure(list(JOB_ROLE = c("Software engineering-related (developer, tester, project manager, architecture)",
"See below", "Student", "Software engineering-related (developer, tester, project manager, architecture)",
"Blogger", "Systems Support", "Student", "IT/Network Administrator",
"Software engineering-related …