R 中的加权频率 Table
Weighted Frequency Table in R
我想在 R 中为某个变量 (INTERVIEW_DAY) 创建一个频率 table，但考虑到另一个变量作为权重 (WEIGHT)。
我已经尝试使用包 data.table 来做到这一点。不过，我很乐意使用 R-Base 包来完成。
下面是我想要的 table 类型，但仍然没有加权，这正是我想要学习的。
数据
（变量 TUCASEID、INTERVIEW_DAY 和 WEIGHT）：
TUCASEID INTERVIEW_DAY WEIGHT
1 2.00301e+13 5 8155462.7
2 2.00301e+13 6 1735322.5
3 2.00301e+13 6 3830527.5
4 2.00301e+13 4 6622023.0
5 2.00301e+13 4 3068387.3
6 2.00301e+13 4 3455424.9
7 2.00301e+13 1 1637826.3
8 2.00301e+13 2 6574426.8
9 2.00301e+13 6 1528296.3
10 2.00301e+13 4 4277052.8
11 2.00301e+13 6 1961482.3
12 2.00301e+13 7 505227.2
13 2.00301e+13 6 2135476.8
14 2.00301e+13 3 5366309.3
15 2.00301e+13 6 1058351.1
使用打包的 data.table 创建 table:
df <- setDT(df)
df_freq_table <- df[,.(Freq = .N), by = INTERVIEW_DAY][, Prop := Freq / sum(Freq)][, Cum := cumsum(100 * Prop / sum(Prop))]
我的输出：
df_freq_table[]
INTERVIEW_DAY Freq Prop Cum
1: 5 1 0.06666667 6.666667
2: 6 6 0.40000000 46.666667
3: 4 4 0.26666667 73.333333
4: 1 1 0.06666667 80.000000
5: 2 1 0.06666667 86.666667
6: 7 1 0.06666667 93.333333
7: 3 1 0.06666667 100.000000
在base R中，我们可以利用xtabs/prop.table。根据 OP 的代码，cumsum 是根据 'INTERVIEW_DAY' 中唯一值的出现顺序计算的。因此，为了避免基于整数值的 sorting，转换为指定 levels 的 factor，通过 [=36= 得到 'WEIGHT' 的 sum ] 使用 xtabs，使用 prop.table 到 return 比例，然后在输出
上应用 cumsum
df$INTERVIEW_DAY <- factor(df$INTERVIEW_DAY, levels = unique(df$INTERVIEW_DAY))
tbl1 <- xtabs(WEIGHT ~ INTERVIEW_DAY, df)
Prop <- prop.table(tbl1)
Cum <- cumsum(100 * Prop / sum(Prop))
Cum
# 5 6 4 1 2 7 3
# 15.71029 39.30705 72.86967 76.02470 88.68935 89.66260 100.00000
out <- data.frame(INTERVIEW_DAY = names(tbl1), Freq = as.numeric(tbl1),
Prop = as.numeric(Prop), Cum = as.numeric(Cum))
row.names(out) <- NULL
out
# INTERVIEW_DAY Freq Prop Cum
#1 5 8155462.7 0.157102906 15.71029
#2 6 12249456.5 0.235967631 39.30705
#3 4 17422888.0 0.335626124 72.86967
#4 1 1637826.3 0.031550297 76.02470
#5 2 6574426.8 0.126646592 88.68935
#6 7 505227.2 0.009732453 89.66260
#7 3 5366309.3 0.103373998 100.00000
如果我们需要加权频率，请使用count
library(dplyr)
df %>%
mutate(INTERVIEW_DAY = factor(INTERVIEW_DAY, levels = unique(INTERVIEW_DAY))) %>%
count(INTERVIEW_DAY, wt = WEIGHT, sort = FALSE) %>%
mutate(Prop = n / sum(n),
Cum = cumsum(100 * Prop/sum(Prop)))
# A tibble: 7 x 4
# INTERVIEW_DAY n Prop Cum
# <fct> <dbl> <dbl> <dbl>
#1 5 8155463. 0.157 15.7
#2 6 12249456. 0.236 39.3
#3 4 17422888 0.336 72.9
#4 1 1637826. 0.0316 76.0
#5 2 6574427. 0.127 88.7
#6 7 505227. 0.00973 89.7
#7 3 5366309. 0.103 100.
或 data.table
library(data.table)
setDT(df)[, .(Freq = sum(WEIGHT)), by = INTERVIEW_DAY
][, Prop := Freq / sum(Freq)][, Cum := cumsum(100 * Prop / sum(Prop))][]
# INTERVIEW_DAY Freq Prop Cum
#1: 5 8155462.7 0.157102906 15.71029
#2: 6 12249456.5 0.235967631 39.30705
#3: 4 17422888.0 0.335626124 72.86967
#4: 1 1637826.3 0.031550297 76.02470
#5: 2 6574426.8 0.126646592 88.68935
#6: 7 505227.2 0.009732453 89.66260
#7: 3 5366309.3 0.103373998 100.00000
数据
df <- structure(list(TUCASEID = c(2.00301e+13, 2.00301e+13, 2.00301e+13,
2.00301e+13, 2.00301e+13, 2.00301e+13, 2.00301e+13, 2.00301e+13,
2.00301e+13, 2.00301e+13, 2.00301e+13, 2.00301e+13, 2.00301e+13,
2.00301e+13, 2.00301e+13), INTERVIEW_DAY = c(5L, 6L, 6L, 4L,
4L, 4L, 1L, 2L, 6L, 4L, 6L, 7L, 6L, 3L, 6L), WEIGHT = c(8155462.7,
1735322.5, 3830527.5, 6622023, 3068387.3, 3455424.9, 1637826.3,
6574426.8, 1528296.3, 4277052.8, 1961482.3, 505227.2, 2135476.8,
5366309.3, 1058351.1)), class = "data.frame", row.names = c("1",
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13",
"14", "15"))
我想在 R 中为某个变量 (INTERVIEW_DAY) 创建一个频率 table，但考虑到另一个变量作为权重 (WEIGHT)。
我已经尝试使用包 data.table 来做到这一点。不过，我很乐意使用 R-Base 包来完成。
下面是我想要的 table 类型，但仍然没有加权，这正是我想要学习的。
数据 （变量 TUCASEID、INTERVIEW_DAY 和 WEIGHT）：
TUCASEID INTERVIEW_DAY WEIGHT
1 2.00301e+13 5 8155462.7
2 2.00301e+13 6 1735322.5
3 2.00301e+13 6 3830527.5
4 2.00301e+13 4 6622023.0
5 2.00301e+13 4 3068387.3
6 2.00301e+13 4 3455424.9
7 2.00301e+13 1 1637826.3
8 2.00301e+13 2 6574426.8
9 2.00301e+13 6 1528296.3
10 2.00301e+13 4 4277052.8
11 2.00301e+13 6 1961482.3
12 2.00301e+13 7 505227.2
13 2.00301e+13 6 2135476.8
14 2.00301e+13 3 5366309.3
15 2.00301e+13 6 1058351.1
使用打包的 data.table 创建 table:
df <- setDT(df)
df_freq_table <- df[,.(Freq = .N), by = INTERVIEW_DAY][, Prop := Freq / sum(Freq)][, Cum := cumsum(100 * Prop / sum(Prop))]
我的输出： df_freq_table[]
INTERVIEW_DAY Freq Prop Cum
1: 5 1 0.06666667 6.666667
2: 6 6 0.40000000 46.666667
3: 4 4 0.26666667 73.333333
4: 1 1 0.06666667 80.000000
5: 2 1 0.06666667 86.666667
6: 7 1 0.06666667 93.333333
7: 3 1 0.06666667 100.000000
在base R中，我们可以利用xtabs/prop.table。根据 OP 的代码，cumsum 是根据 'INTERVIEW_DAY' 中唯一值的出现顺序计算的。因此，为了避免基于整数值的 sorting，转换为指定 levels 的 factor，通过 [=36= 得到 'WEIGHT' 的 sum ] 使用 xtabs，使用 prop.table 到 return 比例，然后在输出
cumsum
df$INTERVIEW_DAY <- factor(df$INTERVIEW_DAY, levels = unique(df$INTERVIEW_DAY))
tbl1 <- xtabs(WEIGHT ~ INTERVIEW_DAY, df)
Prop <- prop.table(tbl1)
Cum <- cumsum(100 * Prop / sum(Prop))
Cum
# 5 6 4 1 2 7 3
# 15.71029 39.30705 72.86967 76.02470 88.68935 89.66260 100.00000
out <- data.frame(INTERVIEW_DAY = names(tbl1), Freq = as.numeric(tbl1),
Prop = as.numeric(Prop), Cum = as.numeric(Cum))
row.names(out) <- NULL
out
# INTERVIEW_DAY Freq Prop Cum
#1 5 8155462.7 0.157102906 15.71029
#2 6 12249456.5 0.235967631 39.30705
#3 4 17422888.0 0.335626124 72.86967
#4 1 1637826.3 0.031550297 76.02470
#5 2 6574426.8 0.126646592 88.68935
#6 7 505227.2 0.009732453 89.66260
#7 3 5366309.3 0.103373998 100.00000
如果我们需要加权频率，请使用count
library(dplyr)
df %>%
mutate(INTERVIEW_DAY = factor(INTERVIEW_DAY, levels = unique(INTERVIEW_DAY))) %>%
count(INTERVIEW_DAY, wt = WEIGHT, sort = FALSE) %>%
mutate(Prop = n / sum(n),
Cum = cumsum(100 * Prop/sum(Prop)))
# A tibble: 7 x 4
# INTERVIEW_DAY n Prop Cum
# <fct> <dbl> <dbl> <dbl>
#1 5 8155463. 0.157 15.7
#2 6 12249456. 0.236 39.3
#3 4 17422888 0.336 72.9
#4 1 1637826. 0.0316 76.0
#5 2 6574427. 0.127 88.7
#6 7 505227. 0.00973 89.7
#7 3 5366309. 0.103 100.
或 data.table
library(data.table)
setDT(df)[, .(Freq = sum(WEIGHT)), by = INTERVIEW_DAY
][, Prop := Freq / sum(Freq)][, Cum := cumsum(100 * Prop / sum(Prop))][]
# INTERVIEW_DAY Freq Prop Cum
#1: 5 8155462.7 0.157102906 15.71029
#2: 6 12249456.5 0.235967631 39.30705
#3: 4 17422888.0 0.335626124 72.86967
#4: 1 1637826.3 0.031550297 76.02470
#5: 2 6574426.8 0.126646592 88.68935
#6: 7 505227.2 0.009732453 89.66260
#7: 3 5366309.3 0.103373998 100.00000
数据
df <- structure(list(TUCASEID = c(2.00301e+13, 2.00301e+13, 2.00301e+13,
2.00301e+13, 2.00301e+13, 2.00301e+13, 2.00301e+13, 2.00301e+13,
2.00301e+13, 2.00301e+13, 2.00301e+13, 2.00301e+13, 2.00301e+13,
2.00301e+13, 2.00301e+13), INTERVIEW_DAY = c(5L, 6L, 6L, 4L,
4L, 4L, 1L, 2L, 6L, 4L, 6L, 7L, 6L, 3L, 6L), WEIGHT = c(8155462.7,
1735322.5, 3830527.5, 6622023, 3068387.3, 3455424.9, 1637826.3,
6574426.8, 1528296.3, 4277052.8, 1961482.3, 505227.2, 2135476.8,
5366309.3, 1058351.1)), class = "data.frame", row.names = c("1",
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13",
"14", "15"))