使用调查数据计算 1% 的富人集中了多少

Question

我想在 R 中计算 1% 的富人集中了多少。例如，也许是 1% 的富人集中了该国总财富的 33%。

我有一个包含变量的调查数据集：

• asset：每个人的总资产值（行）；
• wgt ：与每个个体（行）相关的样本重量。

计算此浓度度量的最佳方法是什么？

非常感谢！

Answer 1

欢迎来到 SO。请不要被反对者气馁，我认为你的问题是完全合理的。假设您的调查设计对象是 scf.design .. 这样，您可以使用 survey of consumer finances 尝试测试用例

# compute the cutoff point for the top percentile
y <- coef( svyquantile( ~ asset , scf.design , 0.99 ) )
# for this, you probably don't need the standard error, 
# so immediately just extract the coefficient (the actual number)
# and discard the standard error term by using the `coef` function
# around the `svyquantile` call

# create a new flag in your survey design object
# that's a 1 if the individual is in the top percentile
# and a 0 for everybody else
scf.design <- update( scf.design , onepct = as.numeric( asset > y ) )

# calculate the aggregate of all assets
# held by the top percentile
# and also held by everybody else
svyby( ~asset , ~onepct , scf.design , svytotal )

# or, if you like, calculate them separately
svytotal( ~asset , subset( scf.design , onepct == 1 ) )

svytotal( ~asset , subset( scf.design , onepct == 0 ) )

# and divide each of those numbers by all assets
# held by everybody
svytotal( ~asset , scf.design )