blocks|key|238458|text|将字典的项传递给DataFrame构造函数，并给出列名。之后，解析Date列以获得Timestamp值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|238459|请注意python+2.x和3.x之间的区别：|238460|在python+2.x中：|238461|df+=+pd.DataFrame(data.items(),+columns=['Date',+'DateValue'])
df['Date']+=+pd.to_datetime(df['Date'])|code-block|syntax|javascript|238462|在Python+3.x中：(需要额外的'list')|238463|df+=+pd.DataFrame(list(data.items()),+columns=['Date',+'DateValue'])
df['Date']+=+pd.to_datetime(df['Date'])|238464|entityMap^0|X|4|15|9|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|10|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|11|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|L|7|13|8|@]|D|@]|E|$M|N]]|$1|S|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|T|$]]

Pass the items of the dictionary to the DataFrame constructor, and give the column names. After that parse the <code>Date</code> column to get <code>Timestamp</code> values.

Note the difference between python 2.x and 3.x:

In python 2.x:

<pre><code>df = pd.DataFrame(data.items(), columns=['Date', 'DateValue'])
df['Date'] = pd.to_datetime(df['Date'])
</code></pre>

In Python 3.x: (requiring an additional 'list')

<pre><code>df = pd.DataFrame(list(data.items()), columns=['Date', 'DateValue'])
df['Date'] = pd.to_datetime(df['Date'])
</code></pre>

blocks|key|22000|text|这里的错误是因为使用标量值调用了DataFrame构造函数(其中的值应该是一个列表/字典/...即具有多个列)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|22001|pd.DataFrame(d)
ValueError:+If+using+all+scalar+values,+you+must+must+pass+an+index|code-block|syntax|javascript|22002|您可以从字典中获取项(即键-值对)：|22003|In+[11]:+pd.DataFrame(d.items())++#+or+list(d.items())+in+python+3
Out[11]:
+++++++++++++0++++1
0+++2012-07-02++392
1+++2012-07-06++392
2+++2012-06-29++391
3+++2012-06-28++391
...

In+[12]:+pd.DataFrame(d.items(),+columns=['Date',+'DateValue'])
Out[12]:
++++++++++Date++DateValue
0+++2012-07-02++++++++392
1+++2012-07-06++++++++392
2+++2012-06-29++++++++391|22004|但我认为传递Series构造函数更有意义：|22005|In+[21]:+s+=+pd.Series(d,+name='DateValue')
Out[21]:
2012-06-08++++388
2012-06-09++++388
2012-06-10++++388

In+[22]:+s.index.name+=+'Date'

In+[23]:+s.reset_index()
Out[23]:
++++++++++Date++DateValue
0+++2012-06-08++++++++388
1+++2012-06-09++++++++388
2+++2012-06-10++++++++388|22006|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The error here, is since calling the DataFrame constructor with scalar values (where it expects values to be a list/dict/... i.e. have multiple columns):

<pre><code>pd.DataFrame(d)
ValueError: If using all scalar values, you must must pass an index
</code></pre>

You could take the items from the dictionary (i.e. the key-value pairs):

<pre><code>In [11]: pd.DataFrame(d.items()) # or list(d.items()) in python 3
Out[11]:
 0 1
0 2012-07-02 392
1 2012-07-06 392
2 2012-06-29 391
3 2012-06-28 391
...

In [12]: pd.DataFrame(d.items(), columns=['Date', 'DateValue'])
Out[12]:
 Date DateValue
0 2012-07-02 392
1 2012-07-06 392
2 2012-06-29 391
</code></pre>

But I think it makes more sense to pass the Series constructor:

<pre><code>In [21]: s = pd.Series(d, name='DateValue')
Out[21]:
2012-06-08 388
2012-06-09 388
2012-06-10 388

In [22]: s.index.name = 'Date'

In [23]: s.reset_index()
Out[23]:
 Date DateValue
0 2012-06-08 388
1 2012-06-09 388
2 2012-06-10 388
</code></pre>

blocks|key|238578|text|接受dict作为参数，并返回一个dataframe，其中dict的键作为索引，值作为列。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|238579|def+dict_to_df(d):
++++df=pd.DataFrame(d.items())
++++df.set_index(0,+inplace=True)
++++return+df|code-block|syntax|javascript|238580|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Accepts a dict as argument and returns a dataframe with the keys of the dict as index and values as a column.

<pre><code>def dict_to_df(d):
 df=pd.DataFrame(d.items())
 df.set_index(0, inplace=True)
 return df
</code></pre>

blocks|key|22053|text|pd.DataFrame({'date'+:+dict_dates.keys()+,+'date_value'+:+dict_dates.values()+})|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|22054|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>pd.DataFrame({'date' : dict_dates.keys() , 'date_value' : dict_dates.values() })
</code></pre>

blocks|key|238622|text|您还可以将字典的键和值传递给新的数据帧，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|238623|import+pandas+as+pd

myDict+=+{<the_dict_from_your_example>]
df+=+pd.DataFrame()
df['Date']+=+myDict.keys()
df['DateValue']+=+myDict.values()|code-block|syntax|javascript|238624|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can also just pass the keys and values of the dictionary to the new dataframe, like so:

<pre><code>import pandas as pd

myDict = {&lt;the_dict_from_your_example&gt;]
df = pd.DataFrame()
df['Date'] = myDict.keys()
df['DateValue'] = myDict.values()
</code></pre>

blocks|key|238661|text|我已经多次遇到这种情况，并且有一个示例字典，它是我从函数get_max_Path()创建的，它返回示例字典：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|238662|{2:+0.3097502930247044,++3:+0.4413177909384636,++4:+0.5197224051562838,++5:+0.5717654946470984,++6:+0.6063959031223476,++7:+0.6365209824708223,++8:+0.655918861281035,++9:+0.680844386645206}|238663|为了将其转换为数据帧，我运行了以下命令：|238664|df+=+pd.DataFrame.from_dict(get_max_path(2),+orient+=+'index').reset_index()|238665|返回一个简单的带有单独索引的两列数据帧：|238666|index++0+0+++2+++0.309750+1+++3+++0.441318|238667|只需使用f.rename(columns={'index':+'Column1',+0:+'Column2'},+inplace=True)重命名这些列|238668|entityMap^0|S|E|0|0|59|0|0|0|24|0|0|0|16|0|4|1U|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|Z|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|13|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|R|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|S|$]]

I have run into this several times and have an example dictionary that I created from a function <code>get_max_Path()</code>, and it returns the sample dictionary:

<code>{2: 0.3097502930247044,
 3: 0.4413177909384636,
 4: 0.5197224051562838,
 5: 0.5717654946470984,
 6: 0.6063959031223476,
 7: 0.6365209824708223,
 8: 0.655918861281035,
 9: 0.680844386645206}</code>

To convert this to a dataframe, I ran the following:

<code>df = pd.DataFrame.from_dict(get_max_path(2), orient = 'index').reset_index()</code>

Returns a simple two column dataframe with a separate index:

<code>index 0
0 2 0.309750
1 3 0.441318</code>

Just rename the columns using <code>f.rename(columns={'index': 'Column1', 0: 'Column2'}, inplace=True)</code>

blocks|key|22155|text|在我的例子中，我希望dict的键和值是DataFrame的列和值。所以唯一对我起作用的是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|22156|data+=+{'adjust_power':+'y',+'af_policy_r_submix_prio_adjust':+'[null]',+'af_rf_info':+'[null]',+'bat_ac':+'3500',+'bat_capacity':+'75'}+

columns+=+list(data.keys())
values+=+list(data.values())
arr_len+=+len(values)

pd.DataFrame(np.array(values,+dtype=object).reshape(1,+arr_len),+columns=columns)|code-block|syntax|javascript|22157|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In my case I wanted keys and values of a dict to be columns and values of DataFrame. So the only thing that worked for me was:

<pre><code>data = {'adjust_power': 'y', 'af_policy_r_submix_prio_adjust': '[null]', 'af_rf_info': '[null]', 'bat_ac': '3500', 'bat_capacity': '75'} 

columns = list(data.keys())
values = list(data.values())
arr_len = len(values)

pd.DataFrame(np.array(values, dtype=object).reshape(1, arr_len), columns=columns)
</code></pre>

blocks|key|22188|text|在将字典转换为pandas数据帧时，您希望关键字是所述数据帧的列，值是行值，您可以简单地将字典括起来，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|22189|>>>+dict_+=+{'key+1':+'value+1',+'key+2':+'value+2',+'key+3':+'value+3'}
>>>+pd.DataFrame([dict_])

++++key+1+++++key+2+++++key+3
0+++value+1+++value+2+++value+3|code-block|syntax|javascript|22190|它为我省去了一些麻烦，所以我希望它能帮助到其他人！|22191|编辑:在the+pandas+docs中，DataFrame构造函数中的data参数的一个选项是字典列表。在这里，我们正在传递一个列表，其中包含一个字典。|offset|length|style|CODE|22192|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.html^0|0|0|0|10|4|4|F|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@$K|10|L|11|M|N]]|9|@$K|12|L|13|1|14]]|A|$]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

When converting a dictionary into a pandas dataframe where you want the keys to be the columns of said dataframe and the values to be the row values, you can do simply put brackets around the dictionary like this:

<pre><code>&gt;&gt;&gt; dict_ = {'key 1': 'value 1', 'key 2': 'value 2', 'key 3': 'value 3'}
&gt;&gt;&gt; pd.DataFrame([dict_])

 key 1 key 2 key 3
0 value 1 value 2 value 3
</code></pre>

It's saved me some headaches so I hope it helps someone out there!

EDIT: In <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.html" rel="noreferrer">the pandas docs</a> one option for the <code>data</code> parameter in the DataFrame constructor is a list of dictionaries. Here we're passing a list with one dictionary in it.

blocks|key|238781|text|我认为您可以在创建字典时对数据格式进行一些更改，然后您可以很容易地将其转换为DataFrame：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|238782|输入：|238783|a={'Dates':['2012-06-08','2012-06-10'],'Date_value':[388,389]}|code-block|syntax|javascript|238784|输出：|238785|{'Date_value':+[388,+389],+'Dates':+['2012-06-08',+'2012-06-10']}|238786|238787|aframe=DataFrame(a)|238788|输出:将是您的DataFrame|238789|你只需要在Sublime或Excel中使用一些文本编辑。|238790|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|X|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Y|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Z|8|@]|9|@]|A|$G|H]]|$1|M|3|C|5|6|7|10|8|@]|9|@]|A|$]]|$1|N|3|O|5|F|7|11|8|@]|9|@]|A|$G|H]]|$1|P|3|Q|5|6|7|12|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|13|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|U|$]]

I think that you can make some changes in your data format when you create dictionary, then you can easily convert it to DataFrame:

input:

<pre><code>a={'Dates':['2012-06-08','2012-06-10'],'Date_value':[388,389]}
</code></pre>

output:

<pre><code>{'Date_value': [388, 389], 'Dates': ['2012-06-08', '2012-06-10']}
</code></pre>

input:

<pre><code>aframe=DataFrame(a)
</code></pre>

output: will be your DataFrame

You just need to use some text editing in somewhere like Sublime or maybe Excel.

blocks|key|238833|text|d+=+{'Date':+list(yourDict.keys()),'Date_Values':+list(yourDict.values())}
df+=+pandas.DataFrame(data=d)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|238834|如果不将yourDict.keys()封装在list()中，那么最终所有的键和值都会被放在每列的每一行中。如下所示：|unstyled|offset|length|style|CODE|238835|Date++\+0+++(2012-06-08,+2012-06-09,+2012-06-10,+2012-06-1...|238836|1+++(2012-06-08,+2012-06-09,+2012-06-10,+2012-06-1...|238837|2+++(2012-06-08,+2012-06-09,+2012-06-10,+2012-06-1...|238838|3+++(2012-06-08,+2012-06-09,+2012-06-10,+2012-06-1...|238839|4+++(2012-06-08,+2012-06-09,+2012-06-10,+2012-06-1...|238840|但是通过添加list()，结果如下所示：|238841|Date++Date_Values+0+++2012-06-08++++++++++388+1+++2012-06-09++++++++++388+2+++2012-06-10++++++++++388+3+++2012-06-11++++++++++389+4+++2012-06-12++++++++++389+...|238842|entityMap^0|0|4|F|M|6|0|0|1P|0|0|1H|0|0|1H|0|0|1H|0|0|1H|0|6|6|0|0|4H|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|11|8|@$G|12|H|13|I|J]|$G|14|H|15|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|16|8|@$G|17|H|18|I|J]]|9|@]|A|$]]|$1|M|3|N|5|F|7|19|8|@$G|1A|H|1B|I|J]]|9|@]|A|$]]|$1|O|3|P|5|F|7|1C|8|@$G|1D|H|1E|I|J]]|9|@]|A|$]]|$1|Q|3|R|5|F|7|1F|8|@$G|1G|H|1H|I|J]]|9|@]|A|$]]|$1|S|3|T|5|F|7|1I|8|@$G|1J|H|1K|I|J]]|9|@]|A|$]]|$1|U|3|V|5|F|7|1L|8|@$G|1M|H|1N|I|J]]|9|@]|A|$]]|$1|W|3|X|5|F|7|1O|8|@$G|1P|H|1Q|I|J]]|9|@]|A|$]]|$1|Y|3|-4|5|F|7|1R|8|@]|9|@]|A|$]]]|Z|$]]

<pre><code>d = {'Date': list(yourDict.keys()),'Date_Values': list(yourDict.values())}
df = pandas.DataFrame(data=d)
</code></pre>

If you don't encapsulate <code>yourDict.keys()</code> inside of <code>list()</code> , then you will end up with all of your keys and values being placed in every row of every column. Like this:

<code>Date \
0 (2012-06-08, 2012-06-09, 2012-06-10, 2012-06-1... 
1 (2012-06-08, 2012-06-09, 2012-06-10, 2012-06-1... 
2 (2012-06-08, 2012-06-09, 2012-06-10, 2012-06-1... 
3 (2012-06-08, 2012-06-09, 2012-06-10, 2012-06-1... 
4 (2012-06-08, 2012-06-09, 2012-06-10, 2012-06-1...</code>

But by adding <code>list()</code> then the result looks like this:

<code>Date Date_Values
0 2012-06-08 388
1 2012-06-09 388
2 2012-06-10 388
3 2012-06-11 389
4 2012-06-12 389
...</code>

blocks|key|238857|text|这就是我的工作原理：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|238858|df=+pd.DataFrame([d.keys(),+d.values()]).T
df.columns=+['keys',+'values']++#+call+them+whatever+you+like|code-block|syntax|javascript|238859|我希望这能帮到你|238860|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is how it worked for me : 

<pre><code>df= pd.DataFrame([d.keys(), d.values()]).T
df.columns= ['keys', 'values'] # call them whatever you like
</code></pre>

I hope this helps

blocks|key|238927|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|238928|📷|atomic|offset|length|238929|238930|附注：特别是，我发现面向行的示例很有帮助；因为记录通常是在外部存储的。|238931|https://pbpython.com/pandas-list-dict.html|238932|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/596900f74837f47ce2538c0ccff5accb.png|imageAlt|1|LINK|MUTABLE|url^0|0|0|1|0|0|0|0|0|16|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@$E|10|F|11|1|12]]|A|$]]|$1|G|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|14|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|15|8|@]|9|@$E|16|F|17|1|18]]|A|$]]|$1|L|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S|T|-4]]|U|$5|V|P|W|A|$X|K]]]]

<a href="https://i.stack.imgur.com/fXP0e.png" rel="noreferrer"><img src="https://i.stack.imgur.com/fXP0e.png" alt="df from lists and dictionaries"></a>

p.s. in particular, I've found Row-Oriented examples helpful; since often that how records are stored externally.

<a href="https://pbpython.com/pandas-list-dict.html" rel="noreferrer">https://pbpython.com/pandas-list-dict.html</a>

blocks|key|22429|text|这就是对我有效的方法，因为我希望有一个单独的索引列|type|unstyled|depth|inlineStyleRanges|entityRanges|data|22430|df+=+pd.DataFrame.from_dict(some_dict,+orient="index").reset_index()
df.columns+=+['A',+'B']|code-block|syntax|javascript|22431|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is what worked for me, since I wanted to have a separate index column

<pre><code>df = pd.DataFrame.from_dict(some_dict, orient="index").reset_index()
df.columns = ['A', 'B']
</code></pre>

blocks|key|22480|text|我找到的最简单的方法是创建一个空的dataframe并追加dict。你需要告诉panda's不要关心索引，否则你会得到错误：TypeError:+Can+only+append+a+dict+if+ignore_index=True|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|22481|import+pandas+as+pd
mydict+=+{'foo':+'bar'}
df+=+pd.DataFrame()
df+=+df.append(mydict,+ignore_index=True)|code-block|syntax|javascript|22482|entityMap^0|1Q|1I|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

The simplest way I found is to create an empty dataframe and append the dict.
You need to tell panda's not to care about the index, otherwise you'll get the error: <code>TypeError: Can only append a dict if ignore_index=True</code>
<pre><code>import pandas as pd
mydict = {'foo': 'bar'}
df = pd.DataFrame()
df = df.append(mydict, ignore_index=True)
</code></pre>

I have a Python dictionary like the following:

<pre><code>{u'2012-06-08': 388,
 u'2012-06-09': 388,
 u'2012-06-10': 388,
 u'2012-06-11': 389,
 u'2012-06-12': 389,
 u'2012-06-13': 389,
 u'2012-06-14': 389,
 u'2012-06-15': 389,
 u'2012-06-16': 389,
 u'2012-06-17': 389,
 u'2012-06-18': 390,
 u'2012-06-19': 390,
 u'2012-06-20': 390,
 u'2012-06-21': 390,
 u'2012-06-22': 390,
 u'2012-06-23': 390,
 u'2012-06-24': 390,
 u'2012-06-25': 391,
 u'2012-06-26': 391,
 u'2012-06-27': 391,
 u'2012-06-28': 391,
 u'2012-06-29': 391,
 u'2012-06-30': 391,
 u'2012-07-01': 391,
 u'2012-07-02': 392,
 u'2012-07-03': 392,
 u'2012-07-04': 392,
 u'2012-07-05': 392,
 u'2012-07-06': 392}
</code></pre>

The keys are <a href="http://en.wikipedia.org/wiki/Unicode" rel="noreferrer">Unicode</a> dates and the values are integers. I would like to convert this into a pandas dataframe by having the dates and their corresponding values as two separate columns. Example: col1: Dates col2: DateValue (the dates are still Unicode and datevalues are still integers)

<pre><code> Date DateValue
0 2012-07-01 391
1 2012-07-02 392
2 2012-07-03 392
. 2012-07-04 392
. ... ...
. ... ...
</code></pre>

Any help in this direction would be much appreciated. I am unable to find resources on the pandas docs to help me with this.

I know one solution might be to convert each key-value pair in this dict, into a dict so the entire structure becomes a dict of dicts, and then we can add each row individually to the dataframe. But I want to know if there is an easier way and a more direct way to do this.

So far I have tried converting the dict into a series object but this doesn't seem to maintain the relationship between the columns:

<pre><code>s = Series(my_dict,index=my_dict.keys())
</code></pre>

Convert Python dict into a dataframe

Python

我有一个Python字典，如下所示：{u'2012-06-08': 388, u'2012-06-09': 388, u'2012-06-10': 388, u'2012-06-11': 389, u'2012-06-12': 389, u'2012-06-13': 389, u'2012-06-14': 389, u...

问将Python字典转换为dataframe
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问将Python字典转换为dataframeEN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问将Python字典转换为dataframe
EN