我正在使用Python 3.8。我有一系列字典，所有字典都有相同的鍵......list_of_dicts = [{"a": 1, "b": 2}, {"a": 1, "b": "zz"}, {"a": 1, "b": "2"}]如何返回所有值都相同的鍵列表？例如，上面的內容只是["a"]因為所有三個字典都有“a”= 1。
6 回答
波斯汪
TA貢獻1811條經驗 獲得超4個贊
array_of_dicts = [{"a": 1, "b": 2}, {"a": 1, "b": "zz", "c": "cc"}, {"a": 1, "b": "2"}]
def get_same_vals(dicts):
keys = []
for key in dicts[0].keys():
is_same = True
for each_dict in array_of_dicts:
if not key in each_dict or each_dict[key] != dicts[0][key]:
is_same = False
if is_same:
keys.append(key)
return keys
print(get_same_vals(array_of_dicts))
繁華開滿天機
TA貢獻1816條經驗 獲得超4個贊
正如其他答案中所建議的，創建一個對每個鍵進行分組的主字典，然后檢查它們的唯一性。
# all keys are the same, so get the list
keys = array_of_dicts[0].keys()
# collapse values into a single dictionary
value_dict = {k: set(d[k] for d in array_of_dicts) for k in keys}
# get list of all single-valued keys
print([k for k, v in value_dict.items() if len(v) == 1])
回首憶惘然
TA貢獻1847條經驗 獲得超11個贊
這是一個可能的解決方案，它也適用于字典結構不同（具有不同/額外鍵）的情況：
array_of_dicts = [{"a": 1, "b": 2}, {"a": 1, "b": "zz"}, {"a": 1, "b": "2"}]
def is_entry_in_all_dicts(key, value):
identical_entries_found = 0
for dict in array_of_dicts:
if key in dict:
if dict[key] == value:
identical_entries_found += 1
if identical_entries_found == len(array_of_dicts):
return True
return False
result = []
for dict in array_of_dicts:
for key, value in dict.items():
if is_entry_in_all_dicts(key, value):
if key not in result:
result.append(key)
print(result)
輸出
['a']
30秒到達戰場
TA貢獻1828條經驗 獲得超6個贊
如果您確定它們都具有相同的鍵，則可以像這樣迭代它們的鍵和列表：
array_of_dicts = [{"a": 1, "b": 2}, {"a": 1, "b": "zz"}, {"a": 1, "b": "2"}]
def get_same_vals(dicts):
keys = []
for key in dicts[0].keys():
is_same = True
for each_dict in dicts:
if each_dict[key] != dicts[0][key]:
is_same = False
if is_same:
keys.append(key)
return keys
print(get_same_vals(array_of_dicts))
# Prints ['a']
對于低效的代碼，我深表歉意；我沒有花那么長時間來編寫這個代碼。
當年話下
TA貢獻1890條經驗 獲得超9個贊
如果每個字典都有相同的鍵，您可以將這些值組合成集合并查找包含一個元素的集合：
[list(x.keys())[0] for x in [{k:set([e[k] for e in list_of_dicts])} for k in list_of_dicts[0]] if len(list(x.values())[0]) == 1]輸出：
['a']
四季花海
TA貢獻1811條經驗 獲得超5個贊
這是一個更簡潔的方法
from functools import reduce
array_of_dicts = [{"a": 1, "b": 2}, {"a": 1, "c": "zz"}, {"a": 1, "d": "2"}]
result = reduce(lambda a, b: a.intersection(b),list(map(lambda x: set(x.keys()),
array_of_dicts)))
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