Flutter Dart-无法将参数类型'String'分配给参数类型'Map <String，dynamic>'

Question

我需要在ChartSampleData()中传递2个参数，但是我遇到了麻烦，因为它说明了这一点

List<ChartSampleData> _list = [];
_list.add(ChartSampleData.fromMap(
                '${formattedDate.toString()}', redeemedToday[index].total));

不能将参数类型'String'分配给参数类型'Map '

这是我的课程

我尝试这样做，

代码：

class ChartSampleData {
  ChartSampleData(this.xValue, this.yValue);
  ChartSampleData.fromMap(
      Map<String, dynamic> dataMap0, Map<String, dynamic> dataMap1)
      : xValue = dataMap0['x'],
        yValue = dataMap1['y'];

  final dynamic xValue;
  final dynamic yValue;
}

Answer 1

您使用了错误的构造函数。使用double代替ChartSampleData。

ChartSampleData.fromMap

Answer 2

我认为您需要像这样更改模型类：

library('data.table')

a = data.table(var1 = c(1,2,3,10,123,12,31,4,6,2), bvar = 1:2)

meansd = function(x, sd = TRUE){
  if(!sd) return(mean(x))
  return(list(mean = mean(x), sd = sd(x)))
  
}

#Presumable syntax:
a[, .(meansd(var1, T), meansd(var1, T)), by = bvar] #data are long on metric which is not ideal
#>    bvar       V1       V2
#> 1:    1     32.8     32.8
#> 2:    1  51.8575  51.8575
#> 3:    2        6        6
#> 4:    2 4.690416 4.690416

#What I was hoping the output would be (dcast call results):
ideal = a[, .(meansd(var1, T), meansd(var1, T)), by = bvar]
ideal[, metric := rep(c('mean', 'sd'),2)]
#ideally the output would look something like this, without the need for the dcast:
dcast(ideal,bvar~metric, value.var = c('V1','V2')) #names are nice, but unimportant
#>    bvar V1_mean    V1_sd V2_mean    V2_sd
#> 1:    1    32.8  51.8575    32.8  51.8575
#> 2:    2       6 4.690416       6 4.690416

您可以使用此website将JSON转换为Dart模型。