R返回NA时，mean函数遇到问题。
这是我的数据（is.numeric返回TRUE）：
[1] 0.0148 0.0033 0.0021 0.0018 0.0013 NA NA NA 0.0075 0.0068 0.0059 0.0037 0.0033 0.0025 0.0019 0.0018 0.0072 0.0064 0.0044
[20] 0.0027 0.0024 0.0024 0.0012 0.0004
总结结果：
summary(Display_df$CTR)
Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
0.000400 0.001900 0.002700 0.003991 0.005900 0.014800 3
平均值的结果：
mean(Display_df$CTR, rm.na = T)
[1] NA
有什么想法吗？我以前从未遇到过这个问题，而且看起来很简单。
答案 0 :(得分：0)
您误会了一个论点。重现 //here is the difference, when you use function, function takes arguements as it is.
const UpdateIcons = (products) => {
// if you use this as component props will be like products props will return the products object again, here you have to use {products} destructuring or products.propducts["One"]
//like this const UpdateIcons = ({products}) => {
if((products["One"] === "A") && (products["Two"] === "B") && (products["Three"] === "C") ) {
return <p>Conditional</p>
}
return <p>Show Always</p>
}
class MyIssue extends Component {
render() {
const { products } = this.props;
return (
<div>
{UpdateIcons(products)}
{/*passing arguements here products will be {"One": "A", "Two": "B"}*/}
<UpdateIcons products={products} />
{/* passing props , props will be same as products */}
</div>)
}
}
class App extends Component {
constructor() {
super();
this.state = {
products: {
"One": "A",
"Two": "B",
"Three": "C"
}
};
}
render() {
return <MyIssue products={this.state.products} />
}
}
的方法如下：
NA这将返回NA，因为R不对NA进行计算。例如test<-c(1,2,3,NA,NA)
mean(test,rm.na=T)
将返回1+NA。正确的参数如下所示
NA是mean(test,na.rm = T)
而不是na.rm。