Python 在for循环中追加字典

我想在for循环中附加dictionary，这样就可以得到一个连接的dictionary。而且，并没有必要所有字典的键都完全相同

对于eq

 one={'a': '2', 'c': 't', 'b': '4'}
 two={'a': '3.4', 'c': '7.6'}
 three={'a': 1.2, 'c': 3.4, 'd': '2.3'}

输出：

combined={'a':['2','3.4','1.2'],'b':'4','c':['t','7.6','3.4'],
                'd':'2.3'}

现在回到原来的问题：

每次for循环迭代时，都会生成一个字典，我想附加它

比如：

 emptydict={}

   for x in z:
      newdict=x.dict()
      emptydict.append(newdict)
      print(emptydict)

---

你可以试试这样的

one = {'a': '2', 'c': 't', 'b': '4'}
two = {'a': '3.4', 'c': '7.6'}
three = {'a': 1.2, 'c': 3.4, 'd': '2.3'}

new_dict = {}
list_dict = [one, two, three]

for d in list_dict:
    for key in d:
        if key not in new_dict:
            new_dict[key] = []
        new_dict[key].append(d[key])

print(new_dict)

---

输出：{'a'：['2'，'3.4'，1.2]，'c'：['t'，'7.6'，3.4]，'b'：['4']，'d'：['2.3']}

您可以尝试类似的方法

one = {'a': '2', 'c': 't', 'b': '4'}
two = {'a': '3.4', 'c': '7.6'}
three = {'a': 1.2, 'c': 3.4, 'd': '2.3'}

new_dict = {}
list_dict = [one, two, three]

for d in list_dict:
    for key in d:
        if key not in new_dict:
            new_dict[key] = []
        new_dict[key].append(d[key])

print(new_dict)

---

输出：{'a'：['2'，'3.4'，1.2]，'c'：['t'，'7.6'，3.4]，'b'：['4']，'d'：['2.3']}

您可以尝试听写理解和列表理解：

new_dict = {k : [j[k] for j in [one,two,three] if k in j] for k in set(list(one.keys())+list(two.keys())+list(three.keys())
# Output : { 'a': ['2', '3.4', 1.2], 'b': ['4'], 'c': ['t', '7.6', 3.4], 'd': ['2.3']}

如果希望列表中不包含仅包含一个元素作为可能值的键，请尝试以下操作：

new_dict =  a = {k : [j[k] for j in [one,two,three] if k in j][0] if len([j[k] for j in [one,two,three] if k in j]) ==1 else [j[k] for j in [one,two,three] if k in j] for k in set(list(one.keys())+list(two.keys())+list(three.keys()))}
# Output : {'a': ['2', '3.4', 1.2], 'b': '4', 'c': ['t', '7.6', 3.4], 'd': '2.3'}

---

您可以尝试听写理解和列表理解：

new_dict = {k : [j[k] for j in [one,two,three] if k in j] for k in set(list(one.keys())+list(two.keys())+list(three.keys())
# Output : { 'a': ['2', '3.4', 1.2], 'b': ['4'], 'c': ['t', '7.6', 3.4], 'd': ['2.3']}

如果希望列表中不包含仅包含一个元素作为可能值的键，请尝试以下操作：

new_dict =  a = {k : [j[k] for j in [one,two,three] if k in j][0] if len([j[k] for j in [one,two,three] if k in j]) ==1 else [j[k] for j in [one,two,three] if k in j] for k in set(list(one.keys())+list(two.keys())+list(three.keys()))}
# Output : {'a': ['2', '3.4', 1.2], 'b': '4', 'c': ['t', '7.6', 3.4], 'd': '2.3'}

试试这个

 one={'a': '2', 'c': 't', 'b': '4'}
 two={'a': '3.4', 'c': '7.6'}
 three={'a': 1.2, 'c': 3.4, 'd': '2.3'}

df = pd.DataFrame([one,two,three])

     a    b    c    d
0    2    4    t  NaN
1  3.4  NaN  7.6  NaN
2  1.2  NaN  3.4  2.3

df.to_dict(orient='list')

输出

试试这个

 one={'a': '2', 'c': 't', 'b': '4'}
 two={'a': '3.4', 'c': '7.6'}
 three={'a': 1.2, 'c': 3.4, 'd': '2.3'}

df = pd.DataFrame([one,two,three])

     a    b    c    d
0    2    4    t  NaN
1  3.4  NaN  7.6  NaN
2  1.2  NaN  3.4  2.3

df.to_dict(orient='list')

输出

---

我已经用了你们的例子-

one = {'a': '2', 'c': 't', 'b': '4'}
two = {'a': '3.4', 'c': '7.6'}
three = {'a': 1.2, 'c': 3.4, 'd': '2.3'}
dicts = [one, two, three]
for dictionary in dicts:
    for key, value in dictionary.items():
        try:
            new[key].append(value)
        except KeyError:
            new[key] = [value]

O/p-

---

我已经用了你们的例子-

one = {'a': '2', 'c': 't', 'b': '4'}
two = {'a': '3.4', 'c': '7.6'}
three = {'a': 1.2, 'c': 3.4, 'd': '2.3'}
dicts = [one, two, three]
for dictionary in dicts:
    for key, value in dictionary.items():
        try:
            new[key].append(value)
        except KeyError:
            new[key] = [value]

O/p-

---

伪代码：必须迭代newdict的槽键并查找和附加到组合的[thiskey]，而不是组合。伪代码：必须迭代newdict的槽键并查找和附加到组合的[thiskey]，而不是组合。