如何在python中分組物品運行

Question

假設我想將分隔小於某個閾值的整數組合在一起。我的具體使用情況是確定的裸露代碼最大塊的測試覆蓋率結果，例如：

groupruns('53, 55, 57, 59, 83, 200, 205, 211, 217, 219, 306, 311, 317, 323, 325, 367, 631, 636, 645, 658, 686, 692, 787, 792, 801, 870, 875, 884, 947, 993, 1134, 1139, 1148, 1158', 3) 
#=> [[53, 55, 57, 59], [83], [200], [205], [211], [217, 219], [306], [311], [317], [323, 325], [367], [631], [636], [645], [658], [686], [692], [787], [792], [801], [870], [875], [884], [947], [993], [1134], [1139], [1148], [1158]] 

Answer 1

itertools.groupby可以幫助我們在這裏。唯一的困難是groupby由一個需要爲每個項目計算的「密鑰」分組，使得每個組由具有相同密鑰的連續項目組成。這意味着我們的keyfunc對象需要保存狀態來執行此任務：

class runner(object): 
    def __init__(self, threshold=1): 
     self.threshold = threshold 
     self.last = None 
     self.key = None 
    def __call__(self,item): 
     if self.last is None: 
      self.last = item 
      self.key = item 
      return item 
     if item - self.last <= self.threshold: 
      self.last = item 
      return self.key 
     else: 
      self.last = item 
      self.key = item 
      return item 

的基本思想是，如果我們的最後一個項目的閾值之內，我們回到目前的關鍵，這是第一項的當前運行。

讓我們看到，在行動：

[list(g) for k,g in itertools.groupby((int(s) for s in '53, 55, 57, 59, 83, 200, 205, 211, 217, 219, 306, 311, 317, 323, 325, 367, 631, 636, 645, 658, 686, 692, 787, 792, 801, 870, 875, 884, 947, 993, 1134, 1139, 1148, 1158'.split(',')), runner(3))] 
#=> [[53, 55, 57, 59], [83], [200], [205], [211], [217, 219], [306], [311], [317], [323, 325], [367], [631], [636], [645], [658], [686], [692], [787], [792], [801], [870], [875], [884], [947], [993], [1134], [1139], [1148], [1158]] 

Answer 2

你可以使用雷蒙德赫廷傑的集羣功能

def cluster(data, maxgap, key=None): 
    """Arrange data into groups where successive elements 
     differ by no more than *maxgap* 

     >>> cluster([1, 6, 9, 100, 102, 105, 109, 134, 139], maxgap=10) 
     [[1, 6, 9], [100, 102, 105, 109], [134, 139]] 

     >>> cluster([1, 6, 9, 99, 100, 102, 105, 134, 139, 141], maxgap=10) 
     [[1, 6, 9], [99, 100, 102, 105], [134, 139, 141]] 

    http://stackoverflow.com/a/14783998/190597 (Raymond Hettinger) 
    """ 
    data.sort() 
    groups = [[data[0]]] 
    for item in data[1:]: 
     if key: 
      val = key(item, groups[-1]) 
     else: 
      val = abs(item - groups[-1][-1]) 
     if val <= maxgap: 
      groups[-1].append(item) 
     else: 
      groups.append([item]) 
    return groups 

data = [53, 55, 57, 59, 83, 200, 205, 211, 217, 219, 306, 311, 317, 323, 325, 367, 631, 636, 645, 658, 686, 692, 787, 792, 801, 870, 875, 884, 947, 993, 1134, 1139, 1148, 1158] 
print(cluster(data, maxgap=3)) 

產量

[[53, 55, 57, 59], [83], [200], [205], [211], [217, 219], [306], [311], [317], [323, 325], [367], [631], [636], [645], [658], [686], [692], [787], [792], [801], [870], [875], [884], [947], [993], [1134], [1139], [1148], [1158]] 

Answer 3

這個怎麼樣，而無需使用任何模塊：

注意：假設它們已經排序。

#!/usr/bin/python 

group = (53, 55, 57, 59, 83, 200, 205, 211, 217, 219, 306, 311, 
     317, 323, 325, 367, 631, 636, 645, 658, 686, 692, 787, 
     792, 801, 870, 875, 884, 947, 993, 1134, 1139, 1148, 
     1158) 

def group_runs(group, step): 
    mark = [0] 
    diff = map(lambda x: (x[1] - x[0]), zip(group[:],group[1:])) 
    [mark.append(i+1) for i,j in enumerate(diff) if j > step] 
    return [list(group[x[0]:x[1]]) for x in zip(mark[::], mark[1::])] 

print group_runs(group, 3) 

輸出：

[[53, 55, 57, 59], [83], [200], [205], [211], [217, 219], [306], [311], [317], [323, 
    325], [367], [631], [636], [645], [658], [686], [692], [787], [792], [801], [870], 
    [875], [884], [947], [993], [1134], [1139], [1148]]