「思维发散」R语言分组计算的N种方法

题: 给定一个数据框，根据某个分组对另一个列进行求值。

举例: 按照年份和月份进行分组，对温度进行求均值，其中温度中存在NA

实现这种操作有很多方法，按照代码长度，列出我想出的解题思路

1. for循环

df$Year_Month <- factor(paste(df$Year, df$Month, sep = "-" ))temp_date <- c()
temp_mean <- c()for (i in unique(df$Year_Month)){tmp_df <- df[df$Year_Month == i,]tmp_mean <- mean(tmp_df$Temperature, na.rm = T)temp_date <- c(temp_date,i)temp_mean <- c(temp_mean, tmp_mean)}res_tmp <- data.frame(date=temp_date, temp_mean=temp_mean)

2. split-lapply-do.call

df$Year_Month <- factor(paste(df$Year, df$Month, sep = "-" ))df_list <- split(df, f=df$Year_Month)temp_mean_list <- lapply(df_list, function(x) mean(x$Temperature, na.rm = T) )
do.call(rbind, temp_mean_list)

3. split-sapply

df$Year_Month <- factor(paste(df$Year, df$Month, sep = "-" ))
df_list <- split(df, f=df$Year_Month)
temp_mean_list <- sapply(df_list, function(x) mean(x$Temperature, na.rm = T) )

4. dplyr

library(dplyr)
df %>% group_by(Year, Month) %>% summarise(temp_mean = mean(Temperature, na.rm = T)) %>% head()

5. SQL

library(sqldf)
sqldf("select avg(Temperature) from df group by Year,Month")

6. aggregate

out <- aggregate(df$Temperature, by=list(df$Year, df$Month), FUN=mean, na.rm=TRUE)

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