Question

我之前已经提出这个问题并收到了一个不符合我意愿的答复。当时我用stata来完成这项工作。但是，由于我经常使用这些数据，我希望使用R来创建我想要的东西。我有按年龄，性别和诊断每日入院的数据集。我希望聚合和重塑数据从长到宽。我怎样才能实现这个目标？样本数据和所需输出如下所示。列标题指定性别，年龄和诊断的前缀。感谢

示例数据

structure(list(diag = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L), .Label = c("card", "cere"), class = "factor"), sex = structure(c(1L, 
1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 
1L, 2L, 2L, 1L, 1L, 2L, 2L), .Label = c("Female", "Male"), class = "factor"), 
    age = structure(c(1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 
    1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L), .Label = c("35-64", 
    "65-74"), class = "factor"), admissions = c(1L, 1L, 0L, 0L, 
    6L, 6L, 6L, 1L, 4L, 0L, 0L, 0L, 4L, 6L, 5L, 2L, 2L, 4L, 1L, 
    0L, 6L, 5L, 6L, 4L), bdate = structure(c(1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 
    3L, 3L, 3L, 3L, 3L), .Label = c("1987-01-01", "1987-01-02", 
    "1987-01-03"), class = "factor")), .Names = c("diag", "sex", 
"age", "admissions", "bdate"), row.names = c(NA, -24L), class = "data.frame")

必需的输出

structure(list(date = structure(1:3, .Label = c("01jan1987", 
"02jan1987", "03jan1987"), class = "factor"), f3564card = c(1L, 
4L, 2L), f6574card = c(1L, 0L, 4L), m3564card = c(0L, 0L, 1L), 
    m6574card = c(0L, 0L, 0L), f3564cere = c(6L, 4L, 6L), f6574cere = c(6L, 
    6L, 5L), m3564cere = c(6L, 5L, 6L), m6574cere = c(1L, 2L, 
    4L)), .Names = c("date", "f3564card", "f6574card", "m3564card", 
"m6574card", "f3564cere", "f6574cere", "m3564cere", "m6574cere"
), class = "data.frame", row.names = c(NA, -3L))

Answer 1

您的数据已经是长格式，可以通过＆＃34; reshape2＆＃34;轻松使用，如下所示：

library(reshape)
dcast(df, bdate ~ sex + age + diag, value.var = "admissions")
#        bdate Female_35-64_card Female_35-64_cere Female_65-74_card Female_65-74_cere
# 1 1987-01-01                 1                 6                 1                 6
# 2 1987-01-02                 4                 4                 0                 6
# 3 1987-01-03                 2                 6                 4                 5
#   Male_35-64_card Male_35-64_cere Male_65-74_card Male_65-74_cere
# 1               0               6               0               1
# 2               0               5               0               2
# 3               1               6               0               4

我没有在您的示例输出中看到任何聚合，但如果需要聚合，您可以使用fun.aggregate中的dcast函数实现此目的。

Answer 2

df <- read.table("D:/Programacao/R/Stackoverflow/Nova pasta/sample.csv",
                                          head = T, dec = '.', sep = ',',
                 stringsAsFactors = F)
head(df)
       date    sex cvd ACS   age
1 01 Jul 91 female   0   0 35-64
2 01 Jul 91   male   0   0 35-64
3 01 Jul 91 female   0   0 35-64
4 01 Jul 91   male   1   1 35-64
5 01 Jul 91 female   0   0 65-74
6 02 Jul 91   male   0   0 65-74

认为cvd和ACS分别与男性和女性不相互排斥，

library(dplyr)
df %.%
  group_by(date, sex, age) %.%
  summarise(vcvd = sum(cvd),
            vacs = sum(ACS))
Source: local data frame [111 x 5]
Groups: date, sex

        date    sex   age vcvd vacs
1  01 Jul 91 female 35-64    0    0
2  01 Jul 91 female 65-74    0    0
3  01 Jul 91   male 35-64    1    1
4  02 Aug 91 female 35-64    0    0
5  02 Jul 91 female 65-74    1    0
6  02 Jul 91   male 65-74    0    0
7  03 Aug 91 female 65-74    0    0
8  03 Jul 91 female 35-64    0    0
9  04 Jul 91   male 35-64    1    0
10 04 Jul 91   male 65-74    0    0
..       ...    ...   ...  ...  ...

聚合并重塑从长到宽

2 个答案: