Question

我想遍历@Override public void onAccessibilityEvent(AccessibilityEvent event) { final String appPackageName = event.getPackageName().toString(); if (appPackageName.equals("testappone")){ AccessibilityNodeInfo source = getRootInActiveWindow(); processEvent(source); } } private void processEvent(AccessibilityNodeInfo source){ StringBuilder sb = new StringBuilder(); processSubEvent(source, 0, sb); String string = sb.toString(); Log.e(TAG, "String " + string); } private void processSubEvent(final AccessibilityNodeInfo source, final int n, final StringBuilder sb){ sb.append(tools.getText(source)); sb.append("\n"); int childCount = source.getChildCount(); for (int i = 0; i < childCount; i++) { AccessibilityNodeInfo child = source.getChild(i); if (child != null) { processSubEvent(child, n + 1, sb); } child.recycle(); } }列表和数据分组模式。

实际上，该列表是具有各种属性的字典的列表，可以将其分为3种类型。我将它们称为python，A和B。

我正在寻找的模式是每个C类型dict以及前一个A dict和前两个C dict。每个B和A字典应仅存在于一组中。

示例：

期望的结果：Original List (data): [A1, B1, B2, B3, C1, A2, A3, B4, B5, B6, B7, C2, B8, C3, A4]

条件：

从示例中可以看到，如果没有先前的[[B2,B3,C1,A2], [B7,B8,C3,A4]]和A（例如B）或存在另一个C，则应忽略A1在A和B（例如C）之前的A3。此外，可能还有流氓C也可以忽略（例如C2）。

我尝试过的事情：

# Extract indices for all A elements
As = [i for i, item in enumerate(data) if item['Class']=="A"]

 # Loop through the A's
for a in As:

    # Ensure the A isn't too close to the start of the list to have sufficient prev elements
    if a > 2:

        e = [data[a]]

        # For each prev item
        for index in range (a-1,0,-1):

            # Get the item
            item = data[index]            

            if (len(e) > 3) :
                continue #Exit once there are 4 items in the list 
            elif (len(e) > 1) :
                searching = "B"; # Start by seraching for B's
            else:
                searching = "C"; # After a B is found go to C's

            if item['Class']=="A": # If another A is found before the list is filled end the search
                break
            elif item['Class']==searching:
                e.append(item)


        if data[index]['Class']=="A":
            continue

这行得通，但是感觉像是很糟糕的代码！任何更好的解决方案建议将不胜感激。

Answer 1

在您的情况下，我会使用Regex

import re

# Convert to Class string representation
# example 'ABBBCAABBBBCBCA' 
string_repr = ''.join([item['Class'] for item in data])

# Compile pattern we are looking for
pattern = re.compile(r'BC*B+C+B*A')

# Find last possition of patterns in string_repr
positions = [match.end() - 1 for match in re.finditer(pattern, string_repr)]

# Use indices from positions on data list. Thay match
your_As = [data[i] for i in positions]

在Python清单中寻找模式

1 个答案: