Ruby 按属性对数组中的连续哈希进行分组

我希望能够从一系列约会中找到nx个连续约会的组

我的数据如下所示：

  [{:event_id=>295, :appointment=>false, :accounted=>false},
   {:event_id=>296, :appointment=>false, :accounted=>false},
   {:event_id=>297, :appointment=>false, :accounted=>false},
   {:event_id=>311, :appointment=>false, :accounted=>false},
   {:event_id=>312, :appointment=>false, :accounted=>false},
   {:event_id=>313, :appointment=>false, :accounted=>false},
   {:event_id=>314, :appointment=>396, :accounted=>false},
   {:event_id=>315, :appointment=>397, :accounted=>false},
   {:event_id=>316, :appointment=>398, :accounted=>false},
   {:event_id=>317, :appointment=>false, :accounted=>false},
   {:event_id=>318, :appointment=>399, :accounted=>false},
   {:event_id=>319, :appointment=>false, :accounted=>false},
   {:event_id=>320, :appointment=>400, :accounted=>false},
   {:event_id=>321, :appointment=>401, :accounted=>false}]>

arr, arr1 = [], []
my_data.each do |e|
    arr << e if e[:appointment] != false
    arr = [] if e[:appointment] == false
    if arr.count == 3
       arr1 += arr
       arr = []
    end
end

2.1.2 :084 > arr1
 => [{:event_id=>314, :appointment=>396, :accounted=>false}, {:event_id=>315, :appointment=>397, :accounted=>false}, {:event_id=>316, :appointment=>398, :accounted=>false}]

我希望能够找到任何3个连续的预约组。因此，在本例中，事件314315316是唯一应该返回的组

老实说，我甚至不知道从哪里开始…

你可以这样尝试：

  [{:event_id=>295, :appointment=>false, :accounted=>false},
   {:event_id=>296, :appointment=>false, :accounted=>false},
   {:event_id=>297, :appointment=>false, :accounted=>false},
   {:event_id=>311, :appointment=>false, :accounted=>false},
   {:event_id=>312, :appointment=>false, :accounted=>false},
   {:event_id=>313, :appointment=>false, :accounted=>false},
   {:event_id=>314, :appointment=>396, :accounted=>false},
   {:event_id=>315, :appointment=>397, :accounted=>false},
   {:event_id=>316, :appointment=>398, :accounted=>false},
   {:event_id=>317, :appointment=>false, :accounted=>false},
   {:event_id=>318, :appointment=>399, :accounted=>false},
   {:event_id=>319, :appointment=>false, :accounted=>false},
   {:event_id=>320, :appointment=>400, :accounted=>false},
   {:event_id=>321, :appointment=>401, :accounted=>false}]>

arr, arr1 = [], []
my_data.each do |e|
    arr << e if e[:appointment] != false
    arr = [] if e[:appointment] == false
    if arr.count == 3
       arr1 += arr
       arr = []
    end
end

2.1.2 :084 > arr1
 => [{:event_id=>314, :appointment=>396, :accounted=>false}, {:event_id=>315, :appointment=>397, :accounted=>false}, {:event_id=>316, :appointment=>398, :accounted=>false}]

arr，arr1=[]，[]
我的U数据。每个都有|
arr-arr1
=>[{：事件id=>314，：约会=>396，：占位=>false}，{：事件id=>315，：约会=>397，：占位=>false}，{：事件id=>316，：约会=>398，：占位=>false}]

您可以这样尝试：

  [{:event_id=>295, :appointment=>false, :accounted=>false},
   {:event_id=>296, :appointment=>false, :accounted=>false},
   {:event_id=>297, :appointment=>false, :accounted=>false},
   {:event_id=>311, :appointment=>false, :accounted=>false},
   {:event_id=>312, :appointment=>false, :accounted=>false},
   {:event_id=>313, :appointment=>false, :accounted=>false},
   {:event_id=>314, :appointment=>396, :accounted=>false},
   {:event_id=>315, :appointment=>397, :accounted=>false},
   {:event_id=>316, :appointment=>398, :accounted=>false},
   {:event_id=>317, :appointment=>false, :accounted=>false},
   {:event_id=>318, :appointment=>399, :accounted=>false},
   {:event_id=>319, :appointment=>false, :accounted=>false},
   {:event_id=>320, :appointment=>400, :accounted=>false},
   {:event_id=>321, :appointment=>401, :accounted=>false}]>

arr, arr1 = [], []
my_data.each do |e|
    arr << e if e[:appointment] != false
    arr = [] if e[:appointment] == false
    if arr.count == 3
       arr1 += arr
       arr = []
    end
end

2.1.2 :084 > arr1
 => [{:event_id=>314, :appointment=>396, :accounted=>false}, {:event_id=>315, :appointment=>397, :accounted=>false}, {:event_id=>316, :appointment=>398, :accounted=>false}]

arr，arr1=[]，[]
我的U数据。每个都有|
arr-arr1
=>[{：事件id=>314，：约会=>396，：占位=>false}，{：事件id=>315，：约会=>397，：占位=>false}，{：事件id=>316，：约会=>398，：占位=>false}]

您可以这样尝试：

  [{:event_id=>295, :appointment=>false, :accounted=>false},
   {:event_id=>296, :appointment=>false, :accounted=>false},
   {:event_id=>297, :appointment=>false, :accounted=>false},
   {:event_id=>311, :appointment=>false, :accounted=>false},
   {:event_id=>312, :appointment=>false, :accounted=>false},
   {:event_id=>313, :appointment=>false, :accounted=>false},
   {:event_id=>314, :appointment=>396, :accounted=>false},
   {:event_id=>315, :appointment=>397, :accounted=>false},
   {:event_id=>316, :appointment=>398, :accounted=>false},
   {:event_id=>317, :appointment=>false, :accounted=>false},
   {:event_id=>318, :appointment=>399, :accounted=>false},
   {:event_id=>319, :appointment=>false, :accounted=>false},
   {:event_id=>320, :appointment=>400, :accounted=>false},
   {:event_id=>321, :appointment=>401, :accounted=>false}]>

arr, arr1 = [], []
my_data.each do |e|
    arr << e if e[:appointment] != false
    arr = [] if e[:appointment] == false
    if arr.count == 3
       arr1 += arr
       arr = []
    end
end

2.1.2 :084 > arr1
 => [{:event_id=>314, :appointment=>396, :accounted=>false}, {:event_id=>315, :appointment=>397, :accounted=>false}, {:event_id=>316, :appointment=>398, :accounted=>false}]

arr，arr1=[]，[]
我的U数据。每个都有|
arr-arr1
=>[{：事件id=>314，：约会=>396，：占位=>false}，{：事件id=>315，：约会=>397，：占位=>false}，{：事件id=>316，：约会=>398，：占位=>false}]

您可以这样尝试：

  [{:event_id=>295, :appointment=>false, :accounted=>false},
   {:event_id=>296, :appointment=>false, :accounted=>false},
   {:event_id=>297, :appointment=>false, :accounted=>false},
   {:event_id=>311, :appointment=>false, :accounted=>false},
   {:event_id=>312, :appointment=>false, :accounted=>false},
   {:event_id=>313, :appointment=>false, :accounted=>false},
   {:event_id=>314, :appointment=>396, :accounted=>false},
   {:event_id=>315, :appointment=>397, :accounted=>false},
   {:event_id=>316, :appointment=>398, :accounted=>false},
   {:event_id=>317, :appointment=>false, :accounted=>false},
   {:event_id=>318, :appointment=>399, :accounted=>false},
   {:event_id=>319, :appointment=>false, :accounted=>false},
   {:event_id=>320, :appointment=>400, :accounted=>false},
   {:event_id=>321, :appointment=>401, :accounted=>false}]>

arr, arr1 = [], []
my_data.each do |e|
    arr << e if e[:appointment] != false
    arr = [] if e[:appointment] == false
    if arr.count == 3
       arr1 += arr
       arr = []
    end
end

2.1.2 :084 > arr1
 => [{:event_id=>314, :appointment=>396, :accounted=>false}, {:event_id=>315, :appointment=>397, :accounted=>false}, {:event_id=>316, :appointment=>398, :accounted=>false}]

arr，arr1=[]，[]
我的U数据。每个都有|
arr-arr1
=>[{：事件id=>314，：约会=>396，：占位=>false}，{：事件id=>315，：约会=>397，：占位=>false}，{：事件id=>316，：约会=>398，：占位=>false}]

您可以使用连续约会对阵列进行分组：

arr = [{:event_id=>295, :appointment=>false, :accounted=>false}, {:event_id=>296, :appointment=>false, :accounted=>false}, {:event_id=>297, :appointment=>false, :accounted=>false}, {:event_id=>311, :appointment=>false, :accounted=>false}, {:event_id=>312, :appointment=>false, :accounted=>false}, {:event_id=>313, :appointment=>false, :accounted=>false}, {:event_id=>314, :appointment=>396, :accounted=>false}, {:event_id=>315, :appointment=>397, :accounted=>false}, {:event_id=>316, :appointment=>398, :accounted=>false}, {:event_id=>317, :appointment=>false, :accounted=>false}, {:event_id=>318, :appointment=>399, :accounted=>false}, {:event_id=>319, :appointment=>false, :accounted=>false}, {:event_id=>320, :appointment=>400, :accounted=>false}, {:event_id=>321, :appointment=>401, :accounted=>false}]

groups = arr.slice_before(appointment: false) { |elt, state|
  a, b = state[:appointment], elt[:appointment] # get previous (a) and current (b) appointment
  state[:appointment] = elt[:appointment]       # update state
  a == false || b == false || a.succ != b       # slicing condition
}

groups.to_a
#=> [[{:event_id=>295, :appointment=>false, :accounted=>false}],
#    [{:event_id=>296, :appointment=>false, :accounted=>false}],
#    [{:event_id=>297, :appointment=>false, :accounted=>false}],
#    [{:event_id=>311, :appointment=>false, :accounted=>false}],
#    [{:event_id=>312, :appointment=>false, :accounted=>false}],
#    [{:event_id=>313, :appointment=>false, :accounted=>false}],
#    [{:event_id=>314, :appointment=>396, :accounted=>false},
#     {:event_id=>315, :appointment=>397, :accounted=>false},
#     {:event_id=>316, :appointment=>398, :accounted=>false}],
#    [{:event_id=>317, :appointment=>false, :accounted=>false}],
#    [{:event_id=>318, :appointment=>399, :accounted=>false}],
#    [{:event_id=>319, :appointment=>false, :accounted=>false}],
#    [{:event_id=>320, :appointment=>400, :accounted=>false},
#     {:event_id=>321, :appointment=>401, :accounted=>false}]]

解释切片条件 在以下情况下，元素将移动到新的组/数组中：

• 以前的约会为false（

  a==false

  ）
• 当前约会为false（

  b==false

  ）
• 上一次约会不是当前约会的继任者（

  a.succ！=b

  ）

否则，即两个约会都不是

false

（我假设约会是

false

或整数），并且前一个约会是当前约会的后续约会，它们保留在同一数组中

查找包含3个或更多项的组现在非常简单：

groups.select { |g| g.size >= 3 }

#=> [[{:event_id=>314, :appointment=>396, :accounted=>false},
#     {:event_id=>315, :appointment=>397, :accounted=>false},
#     {:event_id=>316, :appointment=>398, :accounted=>false}]]

您可以使用按连续约会对阵列进行分组：

arr = [{:event_id=>295, :appointment=>false, :accounted=>false}, {:event_id=>296, :appointment=>false, :accounted=>false}, {:event_id=>297, :appointment=>false, :accounted=>false}, {:event_id=>311, :appointment=>false, :accounted=>false}, {:event_id=>312, :appointment=>false, :accounted=>false}, {:event_id=>313, :appointment=>false, :accounted=>false}, {:event_id=>314, :appointment=>396, :accounted=>false}, {:event_id=>315, :appointment=>397, :accounted=>false}, {:event_id=>316, :appointment=>398, :accounted=>false}, {:event_id=>317, :appointment=>false, :accounted=>false}, {:event_id=>318, :appointment=>399, :accounted=>false}, {:event_id=>319, :appointment=>false, :accounted=>false}, {:event_id=>320, :appointment=>400, :accounted=>false}, {:event_id=>321, :appointment=>401, :accounted=>false}]

groups = arr.slice_before(appointment: false) { |elt, state|
  a, b = state[:appointment], elt[:appointment] # get previous (a) and current (b) appointment
  state[:appointment] = elt[:appointment]       # update state
  a == false || b == false || a.succ != b       # slicing condition
}

groups.to_a
#=> [[{:event_id=>295, :appointment=>false, :accounted=>false}],
#    [{:event_id=>296, :appointment=>false, :accounted=>false}],
#    [{:event_id=>297, :appointment=>false, :accounted=>false}],
#    [{:event_id=>311, :appointment=>false, :accounted=>false}],
#    [{:event_id=>312, :appointment=>false, :accounted=>false}],
#    [{:event_id=>313, :appointment=>false, :accounted=>false}],
#    [{:event_id=>314, :appointment=>396, :accounted=>false},
#     {:event_id=>315, :appointment=>397, :accounted=>false},
#     {:event_id=>316, :appointment=>398, :accounted=>false}],
#    [{:event_id=>317, :appointment=>false, :accounted=>false}],
#    [{:event_id=>318, :appointment=>399, :accounted=>false}],
#    [{:event_id=>319, :appointment=>false, :accounted=>false}],
#    [{:event_id=>320, :appointment=>400, :accounted=>false},
#     {:event_id=>321, :appointment=>401, :accounted=>false}]]

解释切片条件 在以下情况下，元素将移动到新的组/数组中：

• 以前的约会为false（

  a==false

  ）
• 当前约会为false（

  b==false

  ）
• 上一次约会不是当前约会的继任者（

  a.succ！=b

  ）

否则，即两个约会都不是

false

（我假设约会是

false

或整数），并且前一个约会是当前约会的后续约会，它们保留在同一数组中

查找包含3个或更多项的组现在非常简单：

groups.select { |g| g.size >= 3 }

#=> [[{:event_id=>314, :appointment=>396, :accounted=>false},
#     {:event_id=>315, :appointment=>397, :accounted=>false},
#     {:event_id=>316, :appointment=>398, :accounted=>false}]]

您可以使用按连续约会对阵列进行分组：

arr = [{:event_id=>295, :appointment=>false, :accounted=>false}, {:event_id=>296, :appointment=>false, :accounted=>false}, {:event_id=>297, :appointment=>false, :accounted=>false}, {:event_id=>311, :appointment=>false, :accounted=>false}, {:event_id=>312, :appointment=>false, :accounted=>false}, {:event_id=>313, :appointment=>false, :accounted=>false}, {:event_id=>314, :appointment=>396, :accounted=>false}, {:event_id=>315, :appointment=>397, :accounted=>false}, {:event_id=>316, :appointment=>398, :accounted=>false}, {:event_id=>317, :appointment=>false, :accounted=>false}, {:event_id=>318, :appointment=>399, :accounted=>false}, {:event_id=>319, :appointment=>false, :accounted=>false}, {:event_id=>320, :appointment=>400, :accounted=>false}, {:event_id=>321, :appointment=>401, :accounted=>false}]

groups = arr.slice_before(appointment: false) { |elt, state|
  a, b = state[:appointment], elt[:appointment] # get previous (a) and current (b) appointment
  state[:appointment] = elt[:appointment]       # update state
  a == false || b == false || a.succ != b       # slicing condition
}

groups.to_a
#=> [[{:event_id=>295, :appointment=>false, :accounted=>false}],
#    [{:event_id=>296, :appointment=>false, :accounted=>false}],
#    [{:event_id=>297, :appointment=>false, :accounted=>false}],
#    [{:event_id=>311, :appointment=>false, :accounted=>false}],
#    [{:event_id=>312, :appointment=>false, :accounted=>false}],
#    [{:event_id=>313, :appointment=>false, :accounted=>false}],
#    [{:event_id=>314, :appointment=>396, :accounted=>false},
#     {:event_id=>315, :appointment=>397, :accounted=>false},
#     {:event_id=>316, :appointment=>398, :accounted=>false}],
#    [{:event_id=>317, :appointment=>false, :accounted=>false}],
#    [{:event_id=>318, :appointment=>399, :accounted=>false}],
#    [{:event_id=>319, :appointment=>false, :accounted=>false}],
#    [{:event_id=>320, :appointment=>400, :accounted=>false},
#     {:event_id=>321, :appointment=>401, :accounted=>false}]]

解释切片条件 在以下情况下，元素将移动到新的组/数组中：

• 以前的约会为false（

  a==false

  ）
• 当前约会为false（

  b==false

  ）
• 上一次约会不是当前约会的继任者（

  a.succ！=b

  ）

否则，即两个约会都不是

false

（我假设约会是

false

或整数），并且前一个约会是当前约会的后续约会，它们保留在同一数组中

查找包含3个或更多项的组现在非常简单：

groups.select { |g| g.size >= 3 }

#=> [[{:event_id=>314, :appointment=>396, :accounted=>false},
#     {:event_id=>315, :appointment=>397, :accounted=>false},
#     {:event_id=>316, :appointment=>398, :accounted=>false}]]

您可以使用按连续约会对阵列进行分组：

arr = [{:event_id=>295, :appointment=>false, :accounted=>false}, {:event_id=>296, :appointment=>false, :accounted=>false}, {:event_id=>297, :appointment=>false, :accounted=>false}, {:event_id=>311, :appointment=>false, :accounted=>false}, {:event_id=>312, :appointment=>false, :accounted=>false}, {:event_id=>313, :appointment=>false, :accounted=>false}, {:event_id=>314, :appointment=>396, :accounted=>false}, {:event_id=>315, :appointment=>397, :accounted=>false}, {:event_id=>316, :appointment=>398, :accounted=>false}, {:event_id=>317, :appointment=>false, :accounted=>false}, {:event_id=>318, :appointment=>399, :accounted=>false}, {:event_id=>319, :appointment=>false, :accounted=>false}, {:event_id=>320, :appointment=>400, :accounted=>false}, {:event_id=>321, :appointment=>401, :accounted=>false}]

groups = arr.slice_before(appointment: false) { |elt, state|
  a, b = state[:appointment], elt[:appointment] # get previous (a) and current (b) appointment
  state[:appointment] = elt[:appointment]       # update state
  a == false || b == false || a.succ != b       # slicing condition
}

groups.to_a
#=> [[{:event_id=>295, :appointment=>false, :accounted=>false}],
#    [{:event_id=>296, :appointment=>false, :accounted=>false}],
#    [{:event_id=>297, :appointment=>false, :accounted=>false}],
#    [{:event_id=>311, :appointment=>false, :accounted=>false}],
#    [{:event_id=>312, :appointment=>false, :accounted=>false}],
#    [{:event_id=>313, :appointment=>false, :accounted=>false}],
#    [{:event_id=>314, :appointment=>396, :accounted=>false},
#     {:event_id=>315, :appointment=>397, :accounted=>false},
#     {:event_id=>316, :appointment=>398, :accounted=>false}],
#    [{:event_id=>317, :appointment=>false, :accounted=>false}],
#    [{:event_id=>318, :appointment=>399, :accounted=>false}],
#    [{:event_id=>319, :appointment=>false, :accounted=>false}],
#    [{:event_id=>320, :appointment=>400, :accounted=>false},
#     {:event_id=>321, :appointment=>401, :accounted=>false}]]

解释切片条件 在以下情况下，元素将移动到新的组/数组中：

• 以前的约会为false（

  a==false

  ）
• 当前约会为false（

  b==false

  ）
• 上一次约会不是当前约会的继任者（

  a.succ！=b

  ）

否则，即两个约会都不是

false

（我假设约会是

false

或整数），并且前一个约会是当前约会的后续约会，它们保留在同一数组中

查找包含3个或更多项的组现在非常简单：

groups.select { |g| g.size >= 3 }

#=> [[{:event_id=>314, :appointment=>396, :accounted=>false},
#     {:event_id=>315, :appointment=>397, :accounted=>false},
#     {:event_id=>316, :appointment=>398, :accounted=>false}]]

这是我（农业）对它的看法。我还将尝试其他一些解决方案

def remaining_event_apps_consecutive(n)             
        result = []
        consec =[]
        @eas.each do |ea|           
            if ea[:appointment] != false && ea[:accounted] ==false                          
                consec << ea                
            else                
                consec = [] 
            end
            if consec.size == n             
                result << consec
                consec = []
            end
        end
        result
    end

def剩余事件应用程序连续（n）
结果=[]
奉献=[]
@每一个都是
如果ea[：约会]！=false&&ea[：会计]==false
这是我（农业）对它的看法。我还将尝试其他一些解决方案

def remaining_event_apps_consecutive(n)             
        result = []
        consec =[]
        @eas.each do |ea|           
            if ea[:appointment] != false && ea[:accounted] ==false                          
                consec << ea                
            else                
                consec = [] 
            end
            if consec.size == n             
                result << consec
                consec = []
            end
        end
        result
    end

def剩余事件应用程序连续（n）
结果=[]
奉献=[]
@每一个都是
如果ea[：约会]！=false&&ea[：会计]==false
这是我（农业）对它的看法。我还将尝试其他一些解决方案

def remaining_event_apps_consecutive(n)             
        result = []
        consec =[]
        @eas.each do |ea|           
            if ea[:appointment] != false && ea[:accounted] ==false                          
                consec << ea                
            else                
                consec = [] 
            end
            if consec.size == n             
                result << consec
                consec = []
            end
        end
        result
    end

def剩余事件应用程序连续（n）
结果=[]
奉献=[]
@每一个都是
如果ea[：约会]！=false&&ea[：会计]==false
这是我（农业）对它的看法。我还将尝试其他一些解决方案

def remaining_event_apps_consecutive(n)             
        result = []
        consec =[]
        @eas.each do |ea|           
            if ea[:appointment] != false && ea[:accounted] ==false                          
                consec << ea                
            else                
                consec = [] 
            end
            if consec.size == n             
                result << consec
                consec = []
            end
        end
        result
    end

def剩余事件应用程序连续（n）
结果=[]
奉献=[]
@每一个都是
如果ea[：约会]！=false&&ea[：会计]==false
如果事件317丢失，即如果连续约会超过3次，则返回什么？仅前3次连续约会-因为我正在计算哪些约会可以按数量包支付如果