将字符串列表按前n个字母分组

Question

我有一个像这样的字符串集合

["snowy10" "catty20" "manny20" "snowy20" "catty10" "snowy20" "catty30" "manny10" "snowy20"  "manny30"]

希望将其转换为基于字符串的前五个字符分组的集合的集合。

[["snowy10" "snowy20" "snowy20"] ["catty10" "catty20""catty30"]["manny10" ""manny20""  "manny20"]]

在Clojure中寻找解决方案。

Answer 1

该group-by功能在这里很有帮助：

clojure.core/group-by
([f coll])
  Returns a map of the elements of coll keyed by the result of
  f on each element. The value at each key will be a vector of the
  corresponding elements, in the order they appeared in coll.

换句话说，group-by使用给定的函数f为中的每个元素生成一个键coll，并且与该键关联的值是该键的累积元素向量。

在您的示例中，如果我们知道所有输入字符串都必须至少包含5个字符，则可以使用subs。但是使用take以下方法构建更通用的健壮解决方案更加容易：

clojure.core/group-by
([f coll])
  Returns a map of the elements of coll keyed by the result of
  f on each element. The value at each key will be a vector of the
  corresponding elements, in the order they appeared in coll.

给我们：

{(\s \n \o \w \y) ["snowy10" "snowy20" "snowy20" "snowy20"]
 (\c \a \t \t \y) ["catty20" "catty10" "catty30"]
 (\m \a \n \n \y) ["manny20" "manny10" "manny30"]}

这不是我们想要的，我们只想要地图值。为此，我们使用vals：

(def strings ["snowy10" "catty20" "manny20" "snowy20" "catty10" "snowy20" "catty30" "manny10" "snowy20"  "manny30"])

(group-by (partial take 5) strings)

我们得到：

(["snowy10" "snowy20" "snowy20" "snowy20"]
 ["catty20" "catty10" "catty30"]
 ["manny20" "manny10" "manny30"])

更改分组条件就像更改我们提供的“键”功能一样简单group-by。例如，我们可以使用take-last以下命令将每个字符串的最后两个字符分组：

{(\s \n \o \w \y) ["snowy10" "snowy20" "snowy20" "snowy20"]
 (\c \a \t \t \y) ["catty20" "catty10" "catty30"]
 (\m \a \n \n \y) ["manny20" "manny10" "manny30"]}

这使：

(["snowy10" "catty10" "manny10"]
 ["catty20" "manny20" "snowy20" "snowy20" "snowy20"]
 ["catty30" "manny30"])