blocks|key|1733165|text|我认为apply没有访问索引的权限；它将每一行视为一个numpy对象，而不是一个Series，如您所见：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1733166|In+[27]:+s.apply(lambda+x:+type(x))
Out[27]:+
a++b
1++2++++<type+'numpy.float64'>
3++6++++<type+'numpy.float64'>
4++4++++<type+'numpy.float64'>|code-block|syntax|javascript|1733167|要克服这一限制，请将索引提升到列，应用您的函数，并使用原始索引重新创建一个Series。|1733168|Series(s.reset_index().apply(f,+axis=1).values,+index=s.index)|1733169|其他方法可能使用s.get_level_values+(在我看来常常有点难看)，或者使用s.iterrows()，后者可能会慢一些--可能取决于f到底做了什么。|1733170|entityMap^0|3|5|0|0|0|0|8|I|18|C|20|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

I don't believe <code>apply</code> has access to the index; it treats each row as a numpy object, not a Series, as you can see:

<pre><code>In [27]: s.apply(lambda x: type(x))
Out[27]: 
a b
1 2 &lt;type 'numpy.float64'&gt;
3 6 &lt;type 'numpy.float64'&gt;
4 4 &lt;type 'numpy.float64'&gt;
</code></pre>

To get around this limitation, promote the indexes to columns, apply your function, and recreate a Series with the original index.

<pre><code>Series(s.reset_index().apply(f, axis=1).values, index=s.index)
</code></pre>

Other approaches might use <code>s.get_level_values</code>, which often gets a little ugly in my opinion, or <code>s.iterrows()</code>, which is likely to be slower -- perhaps depending on exactly what <code>f</code> does.

blocks|key|592507|text|让它成为一个框架，如果你想要返回标量(所以结果是一个系列)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|592508|设置|592509|In+[11]:+s+=+Series([1,2,3],dtype='float64',index=['a','b','c'])

In+[12]:+s
Out[12]:+
a++++1
b++++2
c++++3
dtype:+float64|code-block|syntax|javascript|592510|打印功能|592511|In+[13]:+def+f(x):
++++print+type(x),+x
++++return+x
+++....:+

In+[14]:+pd.DataFrame(s).apply(f)
<class+'pandas.core.series.Series'>+a++++1
b++++2
c++++3
Name:+0,+dtype:+float64
<class+'pandas.core.series.Series'>+a++++1
b++++2
c++++3
Name:+0,+dtype:+float64
Out[14]:+
+++0
a++1
b++2
c++3|592512|由于您可以在这里返回任何内容，所以只需返回标量(通过name属性访问索引)|offset|length|style|CODE|592513|In+[15]:+pd.DataFrame(s).apply(lambda+x:+5+if+x.name+==+'a'+else+x[0]+,1)
Out[15]:+
a++++5
b++++2
c++++3
dtype:+float64|592514|entityMap^0|0|0|0|0|0|Q|4|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@$O|12|P|13|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Make it a frame, return scalars if you want (so the result is a series)

Setup

<pre><code>In [11]: s = Series([1,2,3],dtype='float64',index=['a','b','c'])

In [12]: s
Out[12]: 
a 1
b 2
c 3
dtype: float64
</code></pre>

Printing function

<pre><code>In [13]: def f(x):
 print type(x), x
 return x
 ....: 

In [14]: pd.DataFrame(s).apply(f)
&lt;class 'pandas.core.series.Series'&gt; a 1
b 2
c 3
Name: 0, dtype: float64
&lt;class 'pandas.core.series.Series'&gt; a 1
b 2
c 3
Name: 0, dtype: float64
Out[14]: 
 0
a 1
b 2
c 3
</code></pre>

Since you can return anything here, just return the scalars (access the index via the <code>name</code> attribute)

<pre><code>In [15]: pd.DataFrame(s).apply(lambda x: 5 if x.name == 'a' else x[0] ,1)
Out[15]: 
a 5
b 2
c 3
dtype: float64
</code></pre>

blocks|key|78943|text|在这里，您可能会发现使用where比使用apply更快：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|78944|In+[11]:+s+=+pd.Series([1.,+2.,+3.],+index=['a'+,'b',+'c'])

In+[12]:+s.where(s.index+!=+'a',+5)
Out[12]:+
a++++5
b++++2
c++++3
dtype:+float64|code-block|syntax|javascript|78945|此外，您还可以对以下任何部分使用numpy样式的逻辑/函数：|78946|In+[13]:+(2+*+s+%2B+1).where((s.index+==+'b')+%7C+(s.index+==+'c'),+-s)
Out[13]:+
a+++-1
b++++5
c++++7
dtype:+float64

In+[14]:+(2+*+s+%2B+1).where(s.index+!=+'a',+-s)
Out[14]:+
a+++-1
b++++5
c++++7
dtype:+float64|78947|我建议测试速度(因为效率相对于应用将取决于功能)。虽然，我发现apply__s更易读.|78948|entityMap^0|C|5|K|5|0|0|0|0|V|5|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

You may find it faster to use <code>where</code> rather than <code>apply</code> here:

<pre><code>In [11]: s = pd.Series([1., 2., 3.], index=['a' ,'b', 'c'])

In [12]: s.where(s.index != 'a', 5)
Out[12]: 
a 5
b 2
c 3
dtype: float64
</code></pre>

Also you can use numpy-style logic/functions to any of the parts:

<pre><code>In [13]: (2 * s + 1).where((s.index == 'b') | (s.index == 'c'), -s)
Out[13]: 
a -1
b 5
c 7
dtype: float64

In [14]: (2 * s + 1).where(s.index != 'a', -s)
Out[14]: 
a -1
b 5
c 7
dtype: float64
</code></pre>

I recommend testing for speed (as efficiency against apply will depend on the function). Although, I find that <code>apply</code>s are more readable...

blocks|key|79707|text|如果使用DataFrame.apply()而不是Series.apply()，则可以将整行作为函数中的参数访问。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|79708|def+f1(row):
++++if+row['I']+<+0.5:
++++++++return+0
++++else:
++++++++return+1

def+f2(row):
++++if+row['N1']==1:
++++++++return+0
++++else:
++++++++return+1

import+pandas+as+pd
import+numpy+as+np
df4+=+pd.DataFrame(np.random.rand(6,1),+columns=list('I'))
df4['N1']=df4.apply(f1,+axis=1)
df4['N2']=df4.apply(f2,+axis=1)|code-block|syntax|javascript|79709|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 access the whole row as argument inside the fucntion if you use DataFrame.apply() instead of Series.apply(). 

<pre><code>def f1(row):
 if row['I'] &lt; 0.5:
 return 0
 else:
 return 1

def f2(row):
 if row['N1']==1:
 return 0
 else:
 return 1

import pandas as pd
import numpy as np
df4 = pd.DataFrame(np.random.rand(6,1), columns=list('I'))
df4['N1']=df4.apply(f1, axis=1)
df4['N2']=df4.apply(f2, axis=1)
</code></pre>

blocks|key|592590|text|转换为DataFrame并沿行应用。您可以以x.name的形式访问索引。x现在也是一个具有1值的Series|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|592591|s.to_frame(0).apply(f,+axis=1)[0]|code-block|syntax|javascript|592592|entityMap^0|3|9|M|6|10|1|1C|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

Convert to <code>DataFrame</code> and apply along row. You can access the index as <code>x.name</code>. <code>x</code> is also a <code>Series</code> now with 1 value

<pre><code>s.to_frame(0).apply(f, axis=1)[0]
</code></pre>

blocks|key|79734|text|使用reset_index()将Series转换为DataFrame，将索引转换为列，然后将apply函数转换为DataFrame。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|79735|棘手的部分是知道reset_index()是如何命名列的，下面是几个例子。|79736|具有单一索引的系列|79737|s=pd.Series({'idx1':+'val1',+'idx2':+'val2'})

def+use_index_and_value(row):
++++return+'I+made+this+with+index+{}+and+value+{}'.format(row['index'],+row[0])

s2+=+s.reset_index().apply(use_index_and_value,+axis=1)

#+The+new+Series+has+an+auto-index;
#+You'll+want+to+replace+that+with+the+index+from+the+original+Series
s2.index+=+s.index
s2|code-block|syntax|javascript|79738|输出：|79739|idx1++++I+made+this+with+index+idx1+and+value+val1
idx2++++I+made+this+with+index+idx2+and+value+val2
dtype:+object|79740|具有多索引系列|79741|这里的概念相同，但您需要以row['level_*']的形式访问索引值，因为这是Series.reset_index()放置索引值的地方。|79742|s=pd.Series({
++++('idx(0,0)',+'idx(0,1)'):+'val1',
++++('idx(1,0)',+'idx(1,1)'):+'val2'
})

def+use_index_and_value(row):
++++return+'made+with+index:+{},{}+&+value:+{}'.format(
++++++++row['level_0'],
++++++++row['level_1'],
++++++++row[0]
++++)

s2+=+s.reset_index().apply(use_index_and_value,+axis=1)

#+Replace+auto+index+with+the+index+from+the+original+Series
s2.index+=+s.index
s2|79743|79744|idx(0,0)++idx(0,1)++++made+with+index:+idx(0,0),idx(0,1)+&+value:+val1
idx(1,0)++idx(1,1)++++made+with+index:+idx(1,0),idx(1,1)+&+value:+val2
dtype:+object|79745|如果您的系列或索引有名称，则需要进行相应的调整。|79746|entityMap^0|2|D|1A|5|0|8|D|0|0|0|0|0|0|D|E|14|K|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1D|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1E|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1F|8|@]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1G|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|6|7|1H|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1I|8|@$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|W|3|X|5|L|7|1N|8|@]|D|@]|E|$M|N]]|$1|Y|3|P|5|6|7|1O|8|@]|D|@]|E|$]]|$1|Z|3|10|5|L|7|1P|8|@]|D|@]|E|$M|N]]|$1|11|3|12|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|13|3|-4|5|6|7|1R|8|@]|D|@]|E|$]]]|14|$]]

Use <code>reset_index()</code> to convert the Series to a DataFrame and the index to a column, and then <code>apply</code> your function to the DataFrame.

The tricky part is knowing how <code>reset_index()</code> names the columns, so here are a couple of examples.

<h2>With a Singly Indexed Series</h2>

<pre><code>s=pd.Series({'idx1': 'val1', 'idx2': 'val2'})

def use_index_and_value(row):
 return 'I made this with index {} and value {}'.format(row['index'], row[0])

s2 = s.reset_index().apply(use_index_and_value, axis=1)

# The new Series has an auto-index;
# You'll want to replace that with the index from the original Series
s2.index = s.index
s2
</code></pre>

Output:

<pre><code>idx1 I made this with index idx1 and value val1
idx2 I made this with index idx2 and value val2
dtype: object
</code></pre>

<h2>With a Multi-Indexed Series</h2>

Same concept here, but you'll need to access the index values as <code>row['level_*']</code> because that's where they're placed by <code>Series.reset_index()</code>.

<pre><code>s=pd.Series({
 ('idx(0,0)', 'idx(0,1)'): 'val1',
 ('idx(1,0)', 'idx(1,1)'): 'val2'
})

def use_index_and_value(row):
 return 'made with index: {},{} &amp; value: {}'.format(
 row['level_0'],
 row['level_1'],
 row[0]
 )

s2 = s.reset_index().apply(use_index_and_value, axis=1)

# Replace auto index with the index from the original Series
s2.index = s.index
s2
</code></pre>

Output:

<pre><code>idx(0,0) idx(0,1) made with index: idx(0,0),idx(0,1) &amp; value: val1
idx(1,0) idx(1,1) made with index: idx(1,0),idx(1,1) &amp; value: val2
dtype: object
</code></pre>

If your series or indexes have names, you will need to adjust accordingly.

blocks|key|1733376|text|Series实现了items()方法，它允许使用列表理解来映射键(即索引值)和值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1733377|给出了一系列：|1733378|In[1]:+seriesA+=+pd.Series([4,+2,+3,+7,+9],+name="A")
In[2]:+seriesA
Out[2]:
0++++4
1++++2
2++++3
3++++7
4++++9
dtype:+int64|code-block|syntax|javascript|1733379|现在，假设函数f接受一个键和一个值：|1733380|def+f(key,+value):
++++return+key+%2B+value|1733381|现在，我们可以通过使用“理解”来创建一个新系列：|1733382|In[1]:+pd.Series(data=[f(k,v)+for+k,+v+in+seriesA.items()],+index=seriesA.index)
Out[1]:
0+++++4
1+++++3
2+++++5
3++++10
4++++13
dtype:+int64|1733383|当然，这并不能利用任何非常规的性能优势，但对于某些操作来说，这是有意义的。|1733384|entityMap^0|0|6|9|7|0|0|0|7|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|19|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1B|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|X|$]]

<code>Series</code> implements the <code>items()</code> method, which enables the use of list comprehensions to map keys (i.e. index values) and values.
Given a series:
<pre><code>In[1]: seriesA = pd.Series([4, 2, 3, 7, 9], name=&quot;A&quot;)
In[2]: seriesA
Out[2]:
0 4
1 2
2 3
3 7
4 9
dtype: int64
</code></pre>
Now, assume function <code>f</code> that takes a key and a value:
<pre><code>def f(key, value):
 return key + value
</code></pre>
We can now create a new series by using a for comprehension:
<pre><code>In[1]: pd.Series(data=[f(k,v) for k, v in seriesA.items()], index=seriesA.index)
Out[1]:
0 4
1 3
2 5
3 10
4 13
dtype: int64
</code></pre>
Of course this doesn't take advantage of any numpy performance goodness, but for some of operations it makes sense.

Lets say I have a MultiIndex Series <code>s</code>:

<pre><code>&gt;&gt;&gt; s
 values
a b
1 2 0.1 
3 6 0.3
4 4 0.7
</code></pre>

and I want to apply a function which uses the index of the row:

<pre><code>def f(x):
 # conditions or computations using the indexes
 if x.index[0] and ...: 
 other = sum(x.index) + ...
 return something
</code></pre>

How can I do <code>s.apply(f)</code> for such a function? What is the recommended way to make this kind of operations? I expect to obtain a new Series with the values resulting from this function applied on each row and the same MultiIndex.

Access index in pandas.Series.apply

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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假设我有一个MultiIndex系列s>>> s     valuesa b1 2  0.1 3 6  0.34 4  0.7我想应用一个函数，它使用行的索引：def f(x):   # conditions or computations using the indexes   if x.index[0] and ....

问pandas.Series.apply中的访问索引
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