我有一个数据显示图表上的各种数据点。
var data = [{name:"s", y:2},{name:"e",y:90},{name:"tt",y:9},{name:"se",y:10}]
现在我知道使用简单数组你可以使用Math.max ....和Math.min来计算最小/最大值，不知道如何用对象数组实现相同的效果？
谢谢！
答案 0 :(得分：1)
您可以使用df = pd.concat(
[df.join(pd.DataFrame(
np.random.randint(10, size=(3, 22)),
columns=list(ascii_uppercase[4:])
))] * 10000, ignore_index=True
)
%%timeit
a = df[['A', 'C']].values
The slowest run took 23.28 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 371 µs per loop
In [305]:
%%timeit
c = ['A', 'C']
p = [df.columns.get_loc(i) for i in c]
a = df.values[:, p].astype(df.dtypes[c[0]])
100 loops, best of 3: 9.62 ms per loop
%timeit np.column_stack([df[col].values for col in ['A', 'C']])
The slowest run took 6.66 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 55.6 µs per loop
获取正确的最小值和最大值。
.map如果您使用的是var data = [{name:"s", y:2},{name:"e",y:90},{name:"tt",y:9},{name:"se",y:10}];
var values = data.map(function( obj ) {
return obj.y;
});
var min = Math.min.apply(null, values);
var max = Math.max.apply(null, values);
，那么您可以摆脱ES6并将其替换为传播运营商。
.apply答案 1 :(得分：0)
您仍然可以使用Math.max和Math.min。首先只需map数据。
const max = Math.max.apply(null, data.map((point) => point.y));
const min = Math.min.apply(null, data.map((point) => point.y));