我有一个 map
调用,它会产生一行计算值,因此我有一个 Array
行,这些行是 Array
任何
,像这样
12-element Array{Array{Any,1},1}:
Any[2015-09-01T00:00:00, 2016-09-01T00:00:00, 98, 53.1]
Any[2015-10-01T00:00:00, 2016-10-01T00:00:00, 92, 58.7]
Any[2015-11-01T00:00:00, 2016-11-01T00:00:00, 130, 64.6]
Any[2015-12-01T00:00:00, 2016-12-01T00:00:00, 135, 67.4]
Any[2016-01-01T00:00:00, 2017-01-01T00:00:00, 206, 59.2]
Any[2016-02-01T00:00:00, 2017-02-01T00:00:00, 246, 54.1]
Any[2016-03-01T00:00:00, 2017-03-01T00:00:00, 254, 53.9]
Any[2016-04-01T00:00:00, 2017-04-01T00:00:00, 268, 65.7]
Any[2016-05-01T00:00:00, 2017-05-01T00:00:00, 265, 61.5]
Any[2016-06-01T00:00:00, 2017-06-01T00:00:00, 303, 52.8]
Any[2016-07-01T00:00:00, 2017-07-01T00:00:00, 301, 59.1]
Any[2016-08-01T00:00:00, 2017-08-01T00:00:00, 273, 54.6]
有没有一种简单的方法可以将其转换为具有列名等的DataFrame
?如果没有简单的方法,我愿意接受更难的方法 :) 我可以想到必须重新运行 map
四次来提取列并构建 DataFrame
从那些,但对于这样一个看似平凡的操作来说,这听起来像是很多代码......
编辑 我可以像这样将行“转换”为列
map(x -> map(y -> y[x], r), collect(1:4)
其中 r
是上面的表格,所以我想一个解决方案是为 DataFrame
构造函数提供列名。因此我的临时解决方案是
DataFrame(map(x -> map(y -> y[x], r), collect(1:4)), [:a, :b, :c, :d])
最佳答案
julia> df
12-element Array{Array{Any,1},1}:
Any["2015-09-01T00:00:00", "2016-09-01T00:00:00", 98, 53.1]
Any["2015-10-01T00:00:00", "2016-10-01T00:00:00", 92, 58.7]
Any["2015-11-01T00:00:00", "2016-11-01T00:00:00", 130, 64.6]
Any["2015-12-01T00:00:00", "2016-12-01T00:00:00", 135, 67.4]
Any["2016-01-01T00:00:00", "2017-01-01T00:00:00", 206, 59.2]
Any["2016-02-01T00:00:00", "2017-02-01T00:00:00", 246, 54.1]
Any["2016-03-01T00:00:00", "2017-03-01T00:00:00", 254, 53.9]
Any["2016-04-01T00:00:00", "2017-04-01T00:00:00", 268, 65.7]
Any["2016-05-01T00:00:00", "2017-05-01T00:00:00", 265, 61.5]
Any["2016-06-01T00:00:00", "2017-06-01T00:00:00", 303, 52.8]
Any["2016-07-01T00:00:00", "2017-07-01T00:00:00", 301, 59.1]
Any["2016-08-01T00:00:00", "2017-08-01T00:00:00", 273, 54.6]
julia> DataFrame(permutedims(Array(DataFrame(map(data,df))), [2, 1]))
12×4 DataFrames.DataFrame
│ Row │ x1 │ x2 │ x3 │ x4 │
├─────┼───────────────────────┼───────────────────────┼─────┼──────┤
│ 1 │ "2015-09-01T00:00:00" │ "2016-09-01T00:00:00" │ 98 │ 53.1 │
│ 2 │ "2015-10-01T00:00:00" │ "2016-10-01T00:00:00" │ 92 │ 58.7 │
│ 3 │ "2015-11-01T00:00:00" │ "2016-11-01T00:00:00" │ 130 │ 64.6 │
│ 4 │ "2015-12-01T00:00:00" │ "2016-12-01T00:00:00" │ 135 │ 67.4 │
│ 5 │ "2016-01-01T00:00:00" │ "2017-01-01T00:00:00" │ 206 │ 59.2 │
│ 6 │ "2016-02-01T00:00:00" │ "2017-02-01T00:00:00" │ 246 │ 54.1 │
│ 7 │ "2016-03-01T00:00:00" │ "2017-03-01T00:00:00" │ 254 │ 53.9 │
│ 8 │ "2016-04-01T00:00:00" │ "2017-04-01T00:00:00" │ 268 │ 65.7 │
│ 9 │ "2016-05-01T00:00:00" │ "2017-05-01T00:00:00" │ 265 │ 61.5 │
│ 10 │ "2016-06-01T00:00:00" │ "2017-06-01T00:00:00" │ 303 │ 52.8 │
│ 11 │ "2016-07-01T00:00:00" │ "2017-07-01T00:00:00" │ 301 │ 59.1 │
│ 12 │ "2016-08-01T00:00:00" │ "2017-08-01T00:00:00" │ 273 │ 54.6 │
我认为您的解决方案要好得多...!
关于dataframe - 如何在 Julia 中将数组的数组转换为 DataFrame?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47290331/