【每日一题】教师资格证考试练习（二）

【解决方案1】：

您可以通过保留对找到的匹配项的引用在线性时间内解决此问题（即不使用双循环）。

我在这里假设每个 C 最多有 2 个匹配项：

dicts = [
{'A':1, 'B':2, 'C':10, 'D':100, 'E':None},
{'A':2, 'B':3, 'C':10, 'D':200, 'E':None},
{'A':3, 'B':4, 'C':20, 'D':300, 'E':None},
{'A':4, 'B':5, 'C':20, 'D':400, 'E':None},
{'A':5, 'B':6, 'C':30, 'D':500, 'E':None},
{'A':6, 'B':7, 'C':30, 'D':600, 'E':None}]
refs = {}
for i, d in enumerate(dicts):
    if d['C'] in refs:    # we already found the first match
        j = refs[d['C']]
        # swap the values
        dicts[j]['E'], dicts[i]['E'] = dicts[i]['A'], dicts[j]['A']
    else:                 # this is the first match
        refs[d['C']] = i  # keep a reference of the index
        # copy A to E in case of a single match
        d['E'] = d['A']
dicts

输出：

[{'A': 1, 'B': 2, 'C': 10, 'D': 100, 'E': 2},
 {'A': 2, 'B': 3, 'C': 10, 'D': 200, 'E': 1},
 {'A': 3, 'B': 4, 'C': 20, 'D': 300, 'E': 4},
 {'A': 4, 'B': 5, 'C': 20, 'D': 400, 'E': 3},
 {'A': 5, 'B': 6, 'C': 30, 'D': 500, 'E': 6},
 {'A': 6, 'B': 7, 'C': 30, 'D': 600, 'E': 5}]

【问题讨论】：

【解决方案2】：

如果“C”总是有 2 个可能的重复项，您只需确保 x 和 y 不同，然后分配“E”：

dicts = [
{'A':1, 'B':2, 'C':10, 'D':100, 'E':None},
{'A':2, 'B':3, 'C':10, 'D':200, 'E':None},
{'A':3, 'B':4, 'C':20, 'D':300, 'E':None},
{'A':4, 'B':5, 'C':20, 'D':400, 'E':None},
{'A':5, 'B':6, 'C':30, 'D':500, 'E':None},
{'A':6, 'B':7, 'C':30, 'D':600, 'E':None}]
for x in dicts:
  for y in dicts:
    if y['C'] == x['C'] and y is not x:
        x['E'] = y['A']
        y['E'] = x['A']
print(dicts)

输出：

[{'A': 1, 'B': 2, 'C': 10, 'D': 100, 'E': 2},
 {'A': 2, 'B': 3, 'C': 10, 'D': 200, 'E': 1},
 {'A': 3, 'B': 4, 'C': 20, 'D': 300, 'E': 4},
 {'A': 4, 'B': 5, 'C': 20, 'D': 400, 'E': 3},
 {'A': 5, 'B': 6, 'C': 30, 'D': 500, 'E': 6},
 {'A': 6, 'B': 7, 'C': 30, 'D': 600, 'E': 5}]

【问题讨论】：

【解决方案3】：

使用 循环您的听写：

import itertools
dicts = [
{'A':1, 'B':2, 'C':10, 'D':100, 'E':None},
{'A':2, 'B':3, 'C':10, 'D':200, 'E':None},
{'A':3, 'B':4, 'C':20, 'D':300, 'E':None},
{'A':4, 'B':5, 'C':20, 'D':400, 'E':None},
{'A':5, 'B':6, 'C':30, 'D':500, 'E':None},
{'A':6, 'B':7, 'C':30, 'D':600, 'E':None}]
for x, y in itertools.combinations(dicts, 2):
    if x['E'] is None: # skip if already set
        if x['C'] == y['C']:
            x['E'] = y['A']
            y['E'] = x['A']

这是有效的，因为在示例 dicts 中只有一对匹配。如果没有，您应该指定如何处理这种情况。

【问题讨论】：

【解决方案4】：

当心 BigO：您的 2 个嵌套 for 循环等于 O(n^2)。

您可以使用哈希表来减少它。

dicts = [
    {"A": 1, "B": 2, "C": 10, "D": 100, "E": None},
    {"A": 2, "B": 3, "C": 10, "D": 200, "E": None},
    {"A": 3, "B": 4, "C": 20, "D": 300, "E": None},
    {"A": 4, "B": 5, "C": 20, "D": 400, "E": None},
    {"A": 5, "B": 6, "C": 30, "D": 500, "E": None},
    {"A": 6, "B": 7, "C": 30, "D": 600, "E": None},
]
c = {}
for i in range(len(dicts)):
    if not c.get(dicts[i]["C"]):
        c.update({dicts[i]["C"]: {"upper_row": i}})
    else:
        c[dicts[i]["C"]].update({"lower_row": i})
print(dicts)
print(c)
for value, duplicates in c.items():
    lower = duplicates.get("lower_row")
    if duplicates.get("lower_row"):
        upper = duplicates.get("upper_row")
        dicts[upper]["E"] = dicts[lower]["A"]
        dicts[lower]["E"] = dicts[upper]["A"]
print(dicts)

输出：

"""
Original state of dicts :
[
{'A': 1, 'B': 2, 'C': 10, 'D': 100, 'E': None},
{'A': 2, 'B': 3, 'C': 10, 'D': 200, 'E': None},
{'A': 3, 'B': 4, 'C': 20, 'D': 300, 'E': None},
{'A': 4, 'B': 5, 'C': 20, 'D': 400, 'E': None},
{'A': 5, 'B': 6, 'C': 30, 'D': 500, 'E': None},
{'A': 6, 'B': 7, 'C': 30, 'D': 600, 'E': None}]
c hashtable :
{10: {'upper_row': 0, 'lower_row': 1}, 20: {'upper_row': 2, 'lower_row': 3}, 30: {'upper_row': 4, 'lower_row': 5}}
Final state of dicts: 
[
{'A': 1, 'B': 2, 'C': 10, 'D': 100, 'E': 2}, 
{'A': 2, 'B': 3, 'C': 10, 'D': 200, 'E': 1},
{'A': 3, 'B': 4, 'C': 20, 'D': 300, 'E': 4},
{'A': 4, 'B': 5, 'C': 20, 'D': 400, 'E': 3},
{'A': 5, 'B': 6, 'C': 30, 'D': 500, 'E': 6},
{'A': 6, 'B': 7, 'C': 30, 'D': 600, 'E': 5}]
"""

【问题讨论】：

【解决方案5】：

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【问题讨论】：