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R:R中有“非字符”命令吗?

data-manipulation r

stats_noob · 技术社区 · 2 年前

我正在使用R编程语言。

我有以下数据集:

factor <- c(1,2,3,4,5,6,7,8,9,10)

var_1 <- as.factor(sample(factor, 10000, replace=TRUE, prob=c(0.1,0.1,0.1,0.1,0.1, 0.1,0.1,0.1,0.1,0.1)))

var_2 <- as.factor(sample(factor, 10000, replace=TRUE, prob=c(0.1,0.1,0.1,0.1,0.1, 0.1,0.1,0.1,0.1,0.1)))

var_3 <- as.factor(sample(factor, 10000, replace=TRUE, prob=c(0.1,0.1,0.1,0.1,0.1, 0.1,0.1,0.1,0.1,0.1)))

var_4 <- as.factor(sample(factor, 10000, replace=TRUE, prob=c(0.1,0.1,0.1,0.1,0.1, 0.1,0.1,0.1,0.1,0.1)))

var_5 <- as.factor(sample(factor, 10000, replace=TRUE, prob=c(0.1,0.1,0.1,0.1,0.1, 0.1,0.1,0.1,0.1,0.1)))

my_data = data.frame(var_1, var_2, var_3, var_4, var_5)

我还有另一个“条件”数据集,将用于查询此数据框:

conditions = data.frame(cond_1 = c("1,3,4", "4,5,6"), cond_2 = c("5,6", "7,8,9"))

我的问题是: 我尝试运行以下命令,根据第一行“条件”从“my_data”中选择行,但这会返回一个空结果:

my_data[my_data$var_1 %in% unlist(conditions[1,1]) &
            my_data$var_2 %in% unlist(conditions[1,2]), ]

[1] var_1 var_2 var_3 var_4 var_5
<0 rows> (or 0-length row.names)

我试图通过“检查”这些条件来进一步了解这一点:

class(conditions[1,1])
[1] "character"

这让我觉得“unlist()”命令不起作用,因为条件本身是“字符”而不是“列表”。

这里是否可以使用与“unlist()”命令相同的等效命令来运行上述语句?

总的来说,我试图产生与我从这段代码中得到的结果相同的结果,但保持了我在上面使用的格式:

my_data[my_data$var_1 %in% c("1", "3", "4") &
            my_data$var_2 %in% c("5", "6"), ]

谢谢

参考: Selecting Rows of Data Based on Multiple Conditions

1 回复 | 直到 2 年前

r2evans 2 年前

在前面, "1,3,4" != 1 .看来你应该用 strsplit(., ",") .

expected <- my_data[my_data$var_1 %in% c("1", "3", "4") & my_data$var_2 %in% c("5", "6"), ]
head(expected)
#     var_1 var_2 var_3 var_4 var_5
# 18      3     6     2     2     9
# 129     3     5     3     2     8
# 133     4     5     6     5     8
# 186     1     6     6    10    10
# 204     4     6     4     2     6
# 207     1     5     3     2     9

out <- my_data[do.call(`&`, 
  Map(`%in%`,
      lapply(my_data[,1:2], as.character), 
      lapply(conditions, function(z) strsplit(z, ",")[[1]]))),]
head(out)
#     var_1 var_2 var_3 var_4 var_5
# 18      3     6     2     2     9
# 129     3     5     3     2     8
# 133     4     5     6     5     8
# 186     1     6     6    10    10
# 204     4     6     4     2     6
# 207     1     5     3     2     9

编辑 :更新以获取新信息 conditions :改变 do.call 到 Reduce :

conditions = data.frame(cond_1 = c("1,3,4", "4,5,6"), cond_2 = c("5,6", "7,8,9"), cond_3 = c("4,6", "9"))
out <- my_data[Reduce(`&`,
  Map(`%in%`,
      lapply(my_data[,1:3], as.character),
      lapply(conditions, function(z) strsplit(z, ",")[[1]]))),]
head(out)
#     var_1 var_2 var_3 var_4 var_5
# 133     4     5     6     5     8
# 186     1     6     6    10    10
# 204     4     6     4     2     6
# 232     1     5     6     5     8
# 332     3     6     6     5    10
# 338     1     5     6     3     6