Pandas：创建一个以列列表作为值的字典
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Pandas: create a dictionary with a list of columns as values
提问by FaCoffee
Given this DataFrame:
鉴于此DataFrame：
import pandas as pd
first=[0,1,2,3,4]
second=[10.2,5.7,7.4,17.1,86.11]
third=['a','b','c','d','e']
fourth=['z','zz','zzz','zzzz','zzzzz']
df=pd.DataFrame({'first':first,'second':second,'third':third,'fourth':fourth})
df=df[['first','second','third','fourth']]
first second third fourth
0 0 10.20 a z
1 1 5.70 b zz
2 2 7.40 c zzz
3 3 17.10 d zzzz
4 4 86.11 e zzzzz
I can create a dictionary out of dfusing
我可以创建一个不df使用的字典
a=df.set_index('first')['second'].to_dict()
so that I can decide what is keysand what is values. But what if you want valuesto be a list of columns, such as secondAND third?
这样我就可以决定什么是keys什么是什么values。但是如果你想values成为一个列的列表，比如secondANDthird呢？
If I try this
如果我试试这个
b=df.set_index('first')[['second','third']].to_dict()
I get a weird dictionary of dictionaries
我得到一本奇怪的字典
{'second': {0: 10.199999999999999,
1: 5.7000000000000002,
2: 7.4000000000000004,
3: 17.100000000000001,
4: 86.109999999999999},
'third': {0: 'a', 1: 'b', 2: 'c', 3: 'd', 4: 'e'}}
Instead, I want a dictionary of lists
相反，我想要一个列表字典
{0: [10.199999999999999,a],
1: [5.7000000000000002,b],
2: [7.4000000000000004,c],
3: [17.100000000000001,d],
4: [86.109999999999999,e]}
How to deal with this?
如何处理？
采纳答案by blacksite
Someone else can probably chime in with a pure-pandas solution, but in a pinch I think this ought to work for you. You'd basically create the dictionary on-the-fly, indexing values in each row instead.
其他人可能会加入纯Pandas解决方案，但在紧要关头，我认为这应该适合你。您基本上是即时创建字典，而不是在每一行中索引值。
d = {df.loc[idx, 'first']: [df.loc[idx, 'second'], df.loc[idx, 'third']] for idx in range(df.shape[0])}
d
Out[5]:
{0: [10.199999999999999, 'a'],
1: [5.7000000000000002, 'b'],
2: [7.4000000000000004, 'c'],
3: [17.100000000000001, 'd'],
4: [86.109999999999999, 'e']}
Edit: You could also do this:
编辑：你也可以这样做：
df['new'] = list(zip(df['second'], df['third']))
df
Out[25]:
first second third fourth new
0 0 10.20 a z (10.2, a)
1 1 5.70 b zz (5.7, b)
2 2 7.40 c zzz (7.4, c)
3 3 17.10 d zzzz (17.1, d)
4 4 86.11 e zzzzz (86.11, e)
df = df[['first', 'new']]
df
Out[27]:
first new
0 0 (10.2, a)
1 1 (5.7, b)
2 2 (7.4, c)
3 3 (17.1, d)
4 4 (86.11, e)
df.set_index('first').to_dict()
Out[28]:
{'new': {0: (10.199999999999999, 'a'),
1: (5.7000000000000002, 'b'),
2: (7.4000000000000004, 'c'),
3: (17.100000000000001, 'd'),
4: (86.109999999999999, 'e')}}
In this approach, you would first create the list (or tuple), you want to keep and then "drop" the other columns. This is basically your original approach, modified.
在这种方法中，您将首先创建要保留的列表（或元组），然后“删除”其他列。这基本上是你原来的方法，修改。
And if you really wanted lists instead of tuples, just mapthe listtype onto that 'new'column:
如果你真的通缉名单，而不是元组，只是map在list类型上是'new'列：
df['new'] = list(map(list, zip(df['second'], df['third'])))
回答by jezrael
回答by EdChum
You can zipthe values:
您可以使用zip以下值：
In [118]:
b=df.set_index('first')[['second','third']].values.tolist()
dict(zip(df['first'].index,b))
Out[118]:
{0: [10.2, 'a'], 1: [5.7, 'b'], 2: [7.4, 'c'], 3: [17.1, 'd'], 4: [86.11, 'e']}
