英文:
how do i perform linear regression on dataframe in julia
问题
M = Matrix(df[:, 2:3])
X = zeros(size(M[:, 1], 1), 2)
X[:, 1] = M[:, 1] # X is the reference data
X[:, 2] .= 1.0
Y = M[:, 2] # Y is the non reference
coeff_pred = X \ Y # this gives you the slope and the intercept
"MethodError: no method matching oneunit(::Type{Any})"
这是我的数据框：
我认为问题可能与矩阵的元素类型为Any有关，但当我尝试将列解析为int64或float64时，出现了不同的错误。
英文:
`
M = Matrix(df[:, 2:3])
X = zeros(size(M[:, 1], 1), 2)
X[:, 1] = M[:, 1] # X is the reference data
X[:, 2] .= 1.0
Y = M[:, 2] # Y is the non reference
coeff_pred = X \ Y # this gives you the slope and the intercept
MethodError: no method matching oneunit(::Type{Any})
this is my dataframe
enter image description here
I think it has to do with the matrix being of type any but when i try parse the columns as int64 or float64 i get a different error
答案1
得分: 2
如果出错，这意味着你的数据框包含非数字值，你需要首先清理它们。
英文:
Do:
M = Matrix{Float64}(df[:, 2:3])
If this errors it means that your data frame contains non-numeric values and you first need to clean them.
答案2
得分: 1
Bogumil的答案很可能是正确的，请考虑：
ERROR: MethodError: no method matching oneunit(::Type{Any})
与
2-element Vector{Float64}:
-3.9999999999999987
4.499999999999999
我会补充一点：当我看到人们的DataFrame仅包含Any类型的列，而没有明显的原因时，90%的情况是他们使用XLSX.jl读取Excel文件，并未使用infer_eltypes关键字。来自文档的说明：
> 使用 infer_eltypes=true 以将数据作为具有类型的向量的Vector{Any}获取。默认值为 infer_eltypes=false。
英文:
Bogumil's answer is likely correct, consider:
julia> Any[1 2; 3 4] \ Any[5, 6]
ERROR: MethodError: no method matching oneunit(::Type{Any})
versus
julia> [1 2; 3 4] \ [5, 6]
2-element Vector{Float64}:
-3.9999999999999987
4.499999999999999
I will add one thing: when I see people with DataFrames consisting of only Any typed columns for no apparent reason, it turns out in 90% of all cases they have been reading in an Excel file using the XLSX.jl and not using the infer_eltypes keyword. From the docs:
> Use infer_eltypes=true to get data as a Vector{Any} of typed vectors. The default value is infer_eltypes=false.
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