频率 table，第二个变量在 R 中为 "analytic weight"

Question

我想在 R 中创建一个频率 table，将另一个变量考虑为权重。

更准确地说，作为 "analytic weight"，例如在 Stata 中。根据其帮助文件，

aweights, or analytic weights, are weights that are inversely
        proportional to the variance of an observation; i.e., the variance of
        the jth observation is assumed to be sigma^2/w_j, where w_j are the
        weights.  Typically, the observations represent averages and the
        weights are the number of elements that gave rise to the average.
        For most Stata commands, the recorded scale of aweights is
        irrelevant; Stata internally rescales them to sum to N, the number of
        observations in your data, when it uses them.

stackflow 成员的宝贵贡献是：

Table_WEIGHT <- xtabs(WEIGHT ~ INTERVIEW_DAY, timeuse_2003)
> Prop <- prop.table(Table_WEIGHT)
> Cum <- cumsum(100 * Prop / sum(Prop))
> Cum
        1         2         3         4         5         6         7 
 14.35397  29.14973  43.23935  57.31355  71.50782  85.80359 100.00000 
> out <- data.frame(INTERVIEW_DAY = names(Table_WEIGHT), Freq = as.numeric(Table_WEIGHT),
+                   Prop = as.numeric(Prop), Cum = as.numeric(Cum))
> out
  INTERVIEW_DAY        Freq      Prop       Cum
1             1 11803438268 0.1435397  14.35397
2             2 12166729888 0.1479576  29.14973
3             3 11586059070 0.1408962  43.23935
4             4 11573379591 0.1407420  57.31355
5             5 11672116808 0.1419427  71.50782
6             6 11755579310 0.1429577  85.80359
7             7 11673877965 0.1419641 100.00000

尽管如此，频率仍然不是我所期望的，因为我们使用的是第二个变量的总和作为权重，而不是上面设置的"analytic weight"。

所需的 table 应该是：

 (mean) |
interview_d |
         ay |      Freq.     Percent        Cum.
------------+-----------------------------------
          1 | 2,974.1424       14.35       14.35
          2 | 3,065.6819       14.80       29.15
          3 | 2,919.3688       14.09       43.24
          4 |2,916.17392       14.07       57.31
          5 |2,941.05299       14.19       71.51
          6 | 2,962.0832       14.30       85.80
          7 | 2,941.4968       14.20      100.00
------------+-----------------------------------
      Total |     20,720      100.00

请注意 "Freq" 是完全不同的。

这里有两个变量(INTERVIEW_DATE)和WEIGHT(WEIGHT)的例子，它们是调查的日期和原始文章中没有指定的权重。

> timeuse_2003$INTERVIEW_DATE[1:15]
 [1] "2003-01-03" "2003-01-04" "2003-01-04" "2003-01-02" "2003-01-09" "2003-01-02" "2003-01-06"
 [8] "2003-01-07" "2003-01-04" "2003-01-09" "2003-01-04" "2003-01-05" "2003-01-04" "2003-01-01"
[15] "2003-01-04"


> timeuse_2003$WEIGHT[1:15]
 [1] 8155462.7 1735322.5 3830527.5 6622023.0 3068387.3 3455424.9 1637826.3 6574426.8 1528296.3
[10] 4277052.8 1961482.3  505227.2 2135476.8 5366309.3 1058351.1

我将感谢任何贡献。

Answer 1

您所要求的可以通过以下方式完成：

library(tidyverse)

a <- tibble(interview_day = 1:7,
            frequency = c(2974.1424, 3065.6819, 2919.3688, 2916.17392, 2941.05299, 2962.0832, 2941.4968)) %>%
  mutate(percent = frequency/sum(frequency),
         cum_pct = cumsum(percent)) %>%
  bind_rows(t(colSums(.)[2:3]) %>% as.data.frame())

这是一个仅使用基础 R 的解决方案：

df <- data.frame(frequency = c(2974.1424, 3065.6819, 2919.3688, 2916.17392, 2941.05299, 2962.0832, 2941.4968))
df$interview_day <- 1:nrow(df)
df$percent <- df$frequency/sum(df$frequency)
df$cum_pct <- cumsum(df$percent)

Answer 2

我根据 Stata 帮助文件找到了一个非优雅的解决方案。我刚刚添加了行

timeuse_2003$N_WEIGHT <- timeuse_2003$WEIGHT * 20720/ sum(timeuse_2003$WEIGHT)

并使用

保留代码

Table_WEIGHT <- xtabs(N_WEIGHT ~ INTERVIEW_DAY, timeuse_2003)
Prop <- prop.table(Table_WEIGHT)
Cum <- cumsum(100 * Prop / sum(Prop))
Cum
Freq_Table <- data.frame(INTERVIEW_DAY = names(Table_WEIGHT), Freq = as.numeric(Table_WEIGHT),
                  Prop = as.numeric(Prop), Cum = as.numeric(Cum))
Freq_Table

然后 table 是正确的，例如：

> Freq_Table
  INTERVIEW_DAY      Freq       Prop        Cum
1             1 2974.1424 0.14353969  14.353969
2             2 3065.6819 0.14795762  29.149731
3             3 2919.3688 0.14089618  43.239349
4             4 2916.1739 0.14074198  57.313547
5             5 2941.0530 0.14194271  71.507819
6             6 2962.0832 0.14295769  85.803587
7             7 2941.4968 0.14196413 100.000000

如果有人能阐明如何用我手动输入的观察次数代替自动输入的次数（此代码将用于不同的数据集，所以我无法更新每一个，每次都切换观察次数。像“.N”这样的就很好了！

谢谢！

频率 table，第二个变量在 R 中为 "analytic weight"

Frequency table with second variable as "analytic weight" in R

r

frequency