Python 从条目具有不同长度的字典创建数据框
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Creating dataframe from a dictionary where entries have different lengths
提问by Josh
Say I have a dictionary with 10 key-value pairs. Each entry holds a numpy array. However, the length of the array is not the same for all of them.
假设我有一个包含 10 个键值对的字典。每个条目都包含一个 numpy 数组。但是，所有数组的长度并不相同。
How can I create a dataframe where each column holds a different entry?
如何创建一个数据框，其中每列都包含不同的条目？
When I try:
当我尝试：
pd.DataFrame(my_dict)
I get:
我得到：
ValueError: arrays must all be the same length
Any way to overcome this? I am happy to have Pandas use NaNto pad those columns for the shorter entries.
有什么办法可以克服这个吗？我很高兴让 PandasNaN为较短的条目填充这些列。
采纳答案by Jeff
In Python 3.x:
在 Python 3.x 中：
In [6]: d = dict( A = np.array([1,2]), B = np.array([1,2,3,4]) )
In [7]: pd.DataFrame(dict([ (k,pd.Series(v)) for k,v in d.items() ]))
Out[7]:
A B
0 1 1
1 2 2
2 NaN 3
3 NaN 4
In Python 2.x:
在 Python 2.x 中：
replace d.items()with d.iteritems().
替换d.items()为d.iteritems().
回答by dezzan
Here's a simple way to do that:
这是一个简单的方法来做到这一点：
In[20]: my_dict = dict( A = np.array([1,2]), B = np.array([1,2,3,4]) )
In[21]: df = pd.DataFrame.from_dict(my_dict, orient='index')
In[22]: df
Out[22]:
0 1 2 3
A 1 2 NaN NaN
B 1 2 3 4
In[23]: df.transpose()
Out[23]:
A B
0 1 1
1 2 2
2 NaN 3
3 NaN 4
回答by user2015487
While this does not directly answer the OP's question. I found this to be an excellent solution for my case when I had unequal arrays and I'd like to share:
虽然这并不能直接回答 OP 的问题。当我有不相等的数组时，我发现这是一个很好的解决方案，我想分享：
In [31]: d = {'one' : Series([1., 2., 3.], index=['a', 'b', 'c']),
....: 'two' : Series([1., 2., 3., 4.], index=['a', 'b', 'c', 'd'])}
....:
In [32]: df = DataFrame(d)
In [33]: df
Out[33]:
one two
a 1 1
b 2 2
c 3 3
d NaN 4
回答by OrangeSherbet
A way of tidying up your syntax, but still do essentially the same thing as these other answers, is below:
一种整理语法但仍然与其他答案基本相同的方法如下：
>>> mydict = {'one': [1,2,3], 2: [4,5,6,7], 3: 8}
>>> dict_df = pd.DataFrame({ key:pd.Series(value) for key, value in mydict.items() })
>>> dict_df
one 2 3
0 1.0 4 8.0
1 2.0 5 NaN
2 3.0 6 NaN
3 NaN 7 NaN
A similar syntax exists for lists, too:
列表也存在类似的语法：
>>> mylist = [ [1,2,3], [4,5], 6 ]
>>> list_df = pd.DataFrame([ pd.Series(value) for value in mylist ])
>>> list_df
0 1 2
0 1.0 2.0 3.0
1 4.0 5.0 NaN
2 6.0 NaN NaN
Another syntax for lists is:
列表的另一种语法是：
>>> mylist = [ [1,2,3], [4,5], 6 ]
>>> list_df = pd.DataFrame({ i:pd.Series(value) for i, value in enumerate(mylist) })
>>> list_df
0 1 2
0 1 4.0 6.0
1 2 5.0 NaN
2 3 NaN NaN
You may additionally have to transpose the result and/or change the column data types (float, integer, etc).
您可能还需要转置结果和/或更改列数据类型（浮点数、整数等）。
回答by jpp
You can also use pd.concatalong axis=1with a list of pd.Seriesobjects:
您还可以与对象列表pd.concat一起axis=1使用pd.Series：
import pandas as pd, numpy as np
d = {'A': np.array([1,2]), 'B': np.array([1,2,3,4])}
res = pd.concat([pd.Series(v, name=k) for k, v in d.items()], axis=1)
print(res)
A B
0 1.0 1
1 2.0 2
2 NaN 3
3 NaN 4
回答by Ismail Hachimi
Both the following lines work perfectly :
以下两行都可以完美运行：
pd.DataFrame.from_dict(df, orient='index').transpose() #A
pd.DataFrame(dict([ (k,pd.Series(v)) for k,v in df.items() ])) #B (Better)
But with %timeit on Jupyter, I've got a ratio of 4x speed for B vs A, which is quite impressive especially when working with a huge data set (mainly with a big number of columns/features).
但是在 Jupyter 上使用 %timeit 时，B 与 A 的速度比为 4 倍，这非常令人印象深刻，尤其是在处理大量数据集（主要是具有大量列/特征）时。
回答by Rohan Chandratre
If you don't want it to show NaNand you have two particular lengths, adding a 'space' in each remaining cell would also work.
如果您不希望它显示NaN并且您有两个特定的长度，则在每个剩余的单元格中添加一个“空格”也可以。
import pandas
long = [6, 4, 7, 3]
short = [5, 6]
for n in range(len(long) - len(short)):
short.append(' ')
df = pd.DataFrame({'A':long, 'B':short}]
# Make sure Excel file exists in the working directory
datatoexcel = pd.ExcelWriter('example1.xlsx',engine = 'xlsxwriter')
df.to_excel(datatoexcel,sheet_name = 'Sheet1')
datatoexcel.save()
A B
0 6 5
1 4 6
2 7
3 3
If you have more than 2 lengths of entries, it is advisable to make a function which uses a similar method.
如果您有超过 2 个长度的条目，建议创建一个使用类似方法的函数。
回答by john joy
pd.DataFrame([my_dict]) will do!
pd.DataFrame([my_dict]) 会做！
