交纏基於Python中

Question

列表對象的屬性我有標記對象的兩個列表對象的兩個列表：

class Marker(): 
    def __init__(self,marker_number, marker_data): 
     self.marker_number = marker_number 
     self.marker_data = marker_data 

，我想做的事情是這樣的：

>>> existing_markers = [Marker(marker_number=1, marker_data= 'a'), 
        Marker(marker_number=2, marker_data= 'b'), 
        Marker(marker_number=3, marker_data= 'c'), 
        Marker(marker_number=4, marker_data= 'd'), 
        Marker(marker_number=5, marker_data= 'e'),] 

>>> new_markers = [ Marker(marker_number=1, marker_data= 'aa'), 
        Marker(marker_number=3, marker_data= 'bb'), 
        Marker(marker_number=5, marker_data= 'cc'),] 

>>> interlace_markers(existing_markers, new_markers) 

[Marker(marker_number=1, marker_data= 'a'), 
Marker(marker_number=2, marker_data= 'aa'), 
Marker(marker_number=3, marker_data= 'b'), 
Marker(marker_number=4, marker_data= 'c'), 
Marker(marker_number=5, marker_data= 'bb'), 
Marker(marker_number=6, marker_data= 'd'), 
Marker(marker_number=7, marker_data= 'e'), 
Marker(marker_number=8, marker_data= 'cc')] 

使當我交錯標記時，new_markers列表中與原始標記列表具有相同編號的標記會出現在原始標記之後，但編號更新會保留編號的序列。這是我目前的做法：

def interlace_markers(current_markers_list, new_markers_list): 

    interlaced_markers = [] 
    #interlace the markers 
    for existing_marker in current_markers_list: 
     interlaced_markers.append(existing_marker) 
     for new_marker in new_markers_list: 
      if new_marker.marker_number== existing_marker.marker_number: 
       interlaced_markers.append(new_marker) 

    #reset the sequence 
    sequence_index = 1 
    for marker in interlaced_markers: 
     marker.marker_number= sequence_index 
     sequence_index += 1 

    return interlaced_markers 

我這裏有兩個問題：

1. 什麼是做到這一點
2. 原始列表將可能是大〜10K最Python的方式，並從創建一個 外部文件，但新的列表將會很短〜100-300長，是 有一個更有效的方式來做到這一點，而不是使用列表？

Answer 1

如果我理解你正確，我認爲這是最容易簡單的排序完成：

interlace_markers = existing_markers + new_markers 
interlace_markers.sort(key = lambda x: x.marker_number) 

現在，如果要更改的編號，你可以做一個簡單的循環：

for i,marker in enumerate(interlace_markers): 
    marker.marker_number = i+1 #indexing is 1 based instead of zero based 

請注意，這是有效的，因爲python的排序是穩定這意味着如果不是必須的順序沒有改變。

這導致O（n）+ O（nlogn）= O（nlogn）算法不是太差。