将字典列表转换为数据帧，其中一列用于键，一列用于值

Question

Let's suppose I have the following list:

list1 = [{'a': 1}, {'b': 2}, {'c': 3}]

Which I want to convert it to a panda dataframe that have two columns: one for the keys, and one for the values.

    keys    values
0    'a'      1
1    'b'      2
2    'c'      3

To do so, I have tried to use pd.DataFrame(list1) and also pd.DataFrame.from_records(list1) , but, in both cases, I get a dataframe like:

     a    b    c
0  1.0  NaN  NaN
1  NaN  2.0  NaN
2  NaN  NaN  3.0

Is there any way to specify what I want? By doing research I could only find the way I am describing above.

Answer 1

Use list comprehension with flattening for list of tuples:

df = pd.DataFrame([(i, j) for a in list1 for i, j in a.items()], 
                   columns=['keys','values'])
print (df)
  keys  values
0    a       1
1    b       2
2    c       3

Detail :

print ([(i, j) for a in list1 for i, j in a.items()])

[('a', 1), ('b', 2), ('c', 3)]

Answer 2

If your keys across dictionaries are unique, you can create a single dictionary and feed to pd.DataFrame.from_dict . This can be facilitated by collections.ChainMap :

from collections import ChainMap

list1 = [{'a': 1}, {'b': 2}, {'c': 3}]

df = pd.DataFrame.from_dict(ChainMap(*list1), orient='index').reset_index()

df.columns = ['key', 'value']

print(df)

  key  value
0   a      1
1   b      2
2   c      3

Alternatively, you can feed directly to pd.DataFrame constructor via a list:

df = pd.DataFrame(list(ChainMap(*list1).items()))

Answer 3

虽然@jezrael完全回答了我的问题，但我想指出，如果你将每个字典（ {'a':1} ）转换为列表（ ['a',1] ），你只需要使用pd.DataFrame(list1) ，获得所需的结果。