【发布时间】:2019-05-31 13:55:56
【问题描述】:
data = data.frame("id"=c(1,2,3,4,5,6,7,8,9,10),
"group"=c(1,1,2,1,2,2,2,2,1,2),
"type"=c(1,1,2,3,2,2,3,3,3,1),
"score1"=c(sample(1:4,10,r=T)),
"score2"=c(sample(1:4,10,r=T)),
"score3"=c(sample(1:4,10,r=T)),
"score4"=c(sample(1:4,10,r=T)),
"score5"=c(sample(1:4,10,r=T)),
"weight1"=c(173,109,136,189,186,146,173,102,178,174),
"weight2"=c(147,187,125,126,120,165,142,129,144,197),
"weight3"=c(103,192,102,159,128,179,195,193,135,145),
"weight4"=c(114,182,199,101,111,116,198,123,119,181),
"weight5"=c(159,125,104,171,166,154,197,124,180,154))
这是我的数据样本。我想要分数变量的人口加权计数,如下所示:
count(data, score1, wt = weight1)
count(data, score2, wt = weight2)
count(data, score3, wt = weight3)
count(data, score4, wt = weight4)
count(data, score5, wt = weight5)
但是我的目标是创建一个类型的循环,这样我就可以为 score1-5 的“组”和“类型”的每个组合执行此操作,并将它们存储在单独的向量中,以便
vec1 = weighted score variable for scores1-5 for group = 1 and type = 1
vec2 = weighted score variable for scores1-5 for group = 1 and type = 2
vec3 = weighted score variable for scores1-5 for group = 1 and type = 3
等等等等。
【问题讨论】:
-
您的预期输出是什么样的?
-
6 个向量(对于组和类型的每个组合),每个向量包含 5 个使用 count() 估计的数字,如示例所示
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最好有一个 set.seed 并且至少对于一个组合来说也期望 oiutput。你展示了
count(data, score1, wt = weight1),这就是我使用的类似逻辑 -
谢谢你,但我为什么要设置一个没有随机的种子?
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@bvowe 我想你有
sample