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julia - 牛虻.jl : How to plot date time based?

转载 作者:行者123 更新时间:2023-12-04 15:34:16 24 4
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我正在尝试根据时间绘制数据:

userId = 2
dateGiven = Datetime.date(2014,2,3)
biometric = "heart-rate"
df = DataFrames.readtable(string("data/user_",userId,"/",dateGiven,"/",biometric,".csv"),
colnames = ["Time", "Heart Rate"],
coltypes = {Int,Int} )

df["Time"] = map((timestamp) -> Datetime.unix2datetime(timestamp, Datetime.UTC) , df["Time"])
plot(df,x="Time",y="Heart Rate",Geom.line)

我有这个错误:
no method *(Float64,DateTime{ISOCalendar,Zone0})
in optimize_ticks at /Users/nhenin/.julia/v0.2/Gadfly/src/ticks.jl:77
in apply_statistic at /Users/nhenin/.julia/v0.2/Gadfly/src/statistics.jl:584
in apply_statistics at /Users/nhenin/.julia/v0.2/Gadfly/src/statistics.jl:35
in render at /Users/nhenin/.julia/v0.2/Gadfly/src/Gadfly.jl:636
in writemime at /Users/nhenin/.julia/v0.2/Gadfly/src/Gadfly.jl:738
in sprint at io.jl:434
in display_dict at /Users/nhenin/.julia/v0.2/IJulia/src/execute_request.jl:35

任何提示?

最佳答案

问题是在函数调用 draw(...) 中将日期时间值转换为来自 DataFrame 的字符串值。 .

看下面的代码

我已经分配了一个变量 dt到日期时间值

julia>dt
100-element Array{DateTime{ISOCalendar,Zone0},1}:
2014-03-21T11:48:10 UTC
2014-03-21T11:48:11 UTC
2014-03-21T11:48:12 UTC
.
.
.
2014-03-21T11:49:47 UTC
2014-03-21T11:49:48 UTC
2014-03-21T11:49:49 UTC

如果您甚至尝试将其转换为字符串,则会出现以下错误
julia> dt[1] = string(dt[1])
no method convert(Type{DateTime{ISOCalendar,Zone0}},ASCIIString)

因此错误
julia> p = plot(Df,x="Time",y="Heart_Rate",Geom.line)
Plot(...)

julia> draw(PNG("HeartRate.png",5inch,5inch),p)
no method *(Float64,DateTime{ISOCalendar,Zone0})

我发现的唯一解决方案是创建一个 String 数组,然后使用新的 String 数组初始化 DataFrame
julia> dt_str = Array(Any,length(dt))
julia> for i=1:length(dt)
dt_str[i] = string(dt[i]);
end

julia> Df = DataFrame(Time = dt_str,Heart_Rate = heart_rate)
100x2 DataFrame:
Time Heart_Rate
[1,] "2014-03-21T11:48:10 UTC" 76.2317
[2,] "2014-03-21T11:48:11 UTC" 80.0101
[3,] "2014-03-21T11:48:12 UTC" 66.6338
.
.
.
[98,] "2014-03-21T11:49:47 UTC" 96.9248
[99,] "2014-03-21T11:49:48 UTC" 94.6423
[100,] "2014-03-21T11:49:49 UTC" 92.9679

现在因为我们只有字符串代替 DataFrame 中的日期时间,所以它不会给出任何类型的错误
julia> p = plot(df,x="Time",y="Heart_Rate",Geom.line)
Plot(...)

julia> draw(PNG("HeartRate.png",5inch,5inch),p)


julia>

enter image description here

关于julia - 牛虻.jl : How to plot date time based?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22536691/

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