将函数应用于 pandas 数据框的每一行以创建两个新列 [英] Apply function to each row of pandas dataframe to create two new columns
问题描述
我有一个熊猫DataFrame，st包含多列:
I have a pandas DataFrame, st containing multiple columns:
<class 'pandas.core.frame.DataFrame'>
DatetimeIndex: 53732 entries, 1993-01-07 12:23:58 to 2012-12-02 20:06:23
Data columns:
Date(dd-mm-yy)_Time(hh-mm-ss) 53732 non-null values
Julian_Day 53732 non-null values
AOT_1020 53716 non-null values
AOT_870 53732 non-null values
AOT_675 53188 non-null values
AOT_500 51687 non-null values
AOT_440 53727 non-null values
AOT_380 51864 non-null values
AOT_340 52852 non-null values
Water(cm) 51687 non-null values
%TripletVar_1020 53710 non-null values
%TripletVar_870 53726 non-null values
%TripletVar_675 53182 non-null values
%TripletVar_500 51683 non-null values
%TripletVar_440 53721 non-null values
%TripletVar_380 51860 non-null values
%TripletVar_340 52846 non-null values
440-870Angstrom 53732 non-null values
380-500Angstrom 52253 non-null values
440-675Angstrom 53732 non-null values
500-870Angstrom 53732 non-null values
340-440Angstrom 53277 non-null values
Last_Processing_Date(dd/mm/yyyy) 53732 non-null values
Solar_Zenith_Angle 53732 non-null values
dtypes: datetime64[ns](1), float64(22), object(1)
我想基于对数据框的每一行应用一个函数为此数据框创建两个新列.我不想多次调用该函数(例如，通过执行两个单独的apply调用)，因为它的计算量很大.我尝试通过两种方式来执行此操作，但它们都不起作用:
I want to create two new columns for this dataframe based on applying a function to each row of the dataframe. I don't want to have to call the function multiple times (eg. by doing two separate apply calls) as it is rather computationally intensive. I have tried doing this in two ways, and neither of them work:
使用apply:
Using apply:
我编写了一个函数，该函数采用Series并返回我想要的值的元组:
I have written a function which takes a Series and returns a tuple of the values I want:
def calculate(s):
a = s['path'] + 2*s['row'] # Simple calc for example
b = s['path'] * 0.153
return (a, b)
尝试将其应用于DataFrame会出现错误:
Trying to apply this to the DataFrame gives an error:
st.apply(calculate, axis=1)
---------------------------------------------------------------------------
AssertionError Traceback (most recent call last)
<ipython-input-248-acb7a44054a7> in <module>()
----> 1 st.apply(calculate, axis=1)
C:\Python27\lib\site-packages\pandas\core\frame.pyc in apply(self, func, axis, broadcast, raw, args, **kwds)
4191 return self._apply_raw(f, axis)
4192 else:
-> 4193 return self._apply_standard(f, axis)
4194 else:
4195 return self._apply_broadcast(f, axis)
C:\Python27\lib\site-packages\pandas\core\frame.pyc in _apply_standard(self, func, axis, ignore_failures)
4274 index = None
4275
-> 4276 result = self._constructor(data=results, index=index)
4277 result.rename(columns=dict(zip(range(len(res_index)), res_index)),
4278 inplace=True)
C:\Python27\lib\site-packages\pandas\core\frame.pyc in __init__(self, data, index, columns, dtype, copy)
390 mgr = self._init_mgr(data, index, columns, dtype=dtype, copy=copy)
391 elif isinstance(data, dict):
--> 392 mgr = self._init_dict(data, index, columns, dtype=dtype)
393 elif isinstance(data, ma.MaskedArray):
394 mask = ma.getmaskarray(data)
C:\Python27\lib\site-packages\pandas\core\frame.pyc in _init_dict(self, data, index, columns, dtype)
521
522 return _arrays_to_mgr(arrays, data_names, index, columns,
--> 523 dtype=dtype)
524
525 def _init_ndarray(self, values, index, columns, dtype=None,
C:\Python27\lib\site-packages\pandas\core\frame.pyc in _arrays_to_mgr(arrays, arr_names, index, columns, dtype)
5411
5412 # consolidate for now
-> 5413 mgr = BlockManager(blocks, axes)
5414 return mgr.consolidate()
5415
C:\Python27\lib\site-packages\pandas\core\internals.pyc in __init__(self, blocks, axes, do_integrity_check)
802
803 if do_integrity_check:
--> 804 self._verify_integrity()
805
806 self._consolidate_check()
C:\Python27\lib\site-packages\pandas\core\internals.pyc in _verify_integrity(self)
892 "items")
893 if block.values.shape[1:] != mgr_shape[1:]:
--> 894 raise AssertionError('Block shape incompatible with manager')
895 tot_items = sum(len(x.items) for x in self.blocks)
896 if len(self.items) != tot_items:
AssertionError: Block shape incompatible with manager
I was then going to assign the values returned from apply to two new columns using the method shown in this question. However, I can't even get to this point! This all works fine if I just return one value.
使用循环:
我首先创建了数据框的两个新列，并将它们设置为None:
I first created two new columns of the dataframe and set them to None:
st['a'] = None
st['b'] = None
然后遍历所有索引，并尝试修改我在其中获得的这些None值，但是我所做的修改似乎没有用.也就是说，没有错误发生，但是DataFrame似乎没有被修改.
Then looped over all of the indices and tried to modify these None values that I'd got in there, but the modifications I did didn't seem to work. That is, no error was generated, but the DataFrame didn't seem to be modified.
for i in st.index:
# do calc here
st.ix[i]['a'] = a
st.ix[i]['b'] = b
我认为这两种方法都行得通，但都没有.那么，我在这里做错了什么?最好的，最"Pythonic"和"pandaonic"方式是什么?
I thought that both of these methods would work, but neither of them did. So, what am I doing wrong here? And what is the best, most 'pythonic' and 'pandaonic' way to do this?
推荐答案
要使第一种方法可行，请尝试返回一个Series而不是一个元组(应用程序抛出异常，因为它不知道如何将行粘回去列数与原始框架不匹配).
To make the first approach work, try returning a Series instead of a tuple (apply is throwing an exception because it doesn't know how to glue the rows back together as the number of columns doesn't match the original frame).
def calculate(s):
a = s['path'] + 2*s['row'] # Simple calc for example
b = s['path'] * 0.153
return pd.Series(dict(col1=a, col2=b))
如果替换，第二种方法应该起作用:
The second approach should work if you replace:
st.ix[i]['a'] = a
具有:
st.ix[i, 'a'] = a
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