Python 从数据框中的列中提取字典值
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Extract dictionary value from column in data frame
提问by michalk
I'm looking for a way to optimize my code.
我正在寻找一种优化代码的方法。
I have entry data in this form:
我有这种形式的条目数据：
import pandas as pn
a=[{'Feature1': 'aa1','Feature2': 'bb1','Feature3': 'cc2' },
{'Feature1': 'aa2','Feature2': 'bb2' },
{'Feature1': 'aa1','Feature2': 'cc1' }
]
b=['num1','num2','num3']
df= pn.DataFrame({'num':b, 'dic':a })
I would like to extract element 'Feature3' from dictionaries in column 'dic'(if exist) in above data frame. So far I was able to solve it but I don't know if this is the fastest way, it seems to be a little bit over complicated.
我想从上述数据框中“dic”列（如果存在）的字典中提取元素“Feature3”。到目前为止，我能够解决它，但我不知道这是否是最快的方法，它似乎有点过于复杂。
Feature3=[]
for idx, row in df['dic'].iteritems():
l=row.keys()
if 'Feature3' in l:
Feature3.append(row['Feature3'])
else:
Feature3.append(None)
df['Feature3']=Feature3
print df
Is there a better/faster/simpler way do extract this Feature3 to separate column in the dataframe?
有没有更好/更快/更简单的方法来提取这个 Feature3 来分隔数据帧中的列？
Thank you in advance for help.
预先感谢您的帮助。
回答by Alexander
You can use a list comprehension to extract feature 3 from each row in your dataframe, returning a list.
您可以使用列表理解从数据框中的每一行中提取特征 3，返回一个列表。
feature3 = [d.get('Feature3') for d in df.dic]
If 'Feature3' is not in dic, it returns None by default.
如果 'Feature3' 不在 中dic，则默认返回 None。
You don't even need pandas, as you can again use a list comprehension to extract the feature from your original dictionary a.
您甚至不需要熊猫，因为您可以再次使用列表理解从原始字典中提取特征a。
feature3 = [d.get('Feature3') for d in a]
回答by as133
df['Feature3'] = df['dic'].apply(lambda x: x.get('Feature3'))
Agree with maxymoo. Consider changing the format of your dataframe.
同意maxymoo。考虑更改数据框的格式。
(Sidenote: pandas is generally imported as pd)
（旁注：pandas 通常作为 pd 导入）
回答by Ami Tavory
If you applya Series, you get a quite nice DataFrame:
如果你apply是Series，你会得到一个很好的DataFrame：
>>> df.dic.apply(pn.Series)
Feature1 Feature2 Feature3
0 aa1 bb1 cc2
1 aa2 bb2 NaN
2 aa1 cc1 NaN
From this point, you can just use regular pandas operations.
从这一点来看，您可以只使用常规的 Pandas 操作。
回答by maxymoo
I think you're thinking about the data structures slightly wrong. It's better to create the data frame with the features as columns from the start; pandas is actually smart enough to do this by default:
我认为您正在考虑的数据结构略有错误。最好从一开始就创建以特征为列的数据框；pandas 实际上很聪明，可以默认执行此操作：
In [240]: pd.DataFrame(a)
Out[240]:
Feature1 Feature2 Feature3
0 aa1 bb1 cc2
1 aa2 bb2 NaN
2 aa1 cc1 NaN
You would then add on your "num" column in a separate step, since the data is in a different orientation, either with
然后，您将在单独的步骤中添加“num”列，因为数据处于不同的方向，要么使用
df['num'] = b
or
或者
df = df.assign(num = b)
(I prefer the second option since it's got a more functional flavour).
（我更喜欢第二种选择，因为它具有更实用的风味）。
回答by jezrael
I think you can first create new DataFrameby comprehensionand then create new column like:
我认为您可以先创建 new DataFramebycomprehension然后创建新列，例如：
df1 = pd.DataFrame([x for x in df['dic']])
print df1
Feature1 Feature2 Feature3
0 aa1 bb1 cc2
1 aa2 bb2 NaN
2 aa1 cc1 NaN
df['Feature3'] = df1['Feature3']
print df
dic num Feature3
0 {u'Feature2': u'bb1', u'Feature3': u'cc2', u'F... num1 cc2
1 {u'Feature2': u'bb2', u'Feature1': u'aa2'} num2 NaN
2 {u'Feature2': u'cc1', u'Feature1': u'aa1'} num3 NaN
Or one line:
或一行：
df['Feature3'] = pd.DataFrame([x for x in df['dic']])['Feature3']
print df
dic num Feature3
0 {u'Feature2': u'bb1', u'Feature3': u'cc2', u'F... num1 cc2
1 {u'Feature2': u'bb2', u'Feature1': u'aa2'} num2 NaN
2 {u'Feature2': u'cc1', u'Feature1': u'aa1'} num3 NaN
Timings:
时间：
len(df) = 3:
len(df) = 3：
In [24]: %timeit pd.DataFrame([x for x in df['dic']])
The slowest run took 4.63 times longer than the fastest. This could mean that an intermediate result is being cached
1000 loops, best of 3: 596 μs per loop
In [25]: %timeit df.dic.apply(pn.Series)
1000 loops, best of 3: 1.43 ms per loop
len(df) = 3000:
len(df) = 3000：
In [27]: %timeit pd.DataFrame([x for x in df['dic']])
100 loops, best of 3: 3.16 ms per loop
In [28]: %timeit df.dic.apply(pn.Series)
1 loops, best of 3: 748 ms per loop
