blocks|key|141854|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|141855|g+=+pd.DataFrame(['A','B','A','C','D','D','E'])

#+Group+by+the+contents+of+column+0+
gg+=+g.groupby(0)++

#+Create+a+DataFrame+with+the+counts+of+each+letter
histo+=+gg.apply(lambda+x:+x.count())

#+Add+a+new+column+that+is+the+count+/+total+number+of+elements++++
histo[1]+=+histo.astype(np.float)/len(g)+

print+histo|code-block|syntax|javascript|141856|输出：|141857|+++0+++++++++1
0+++++++++++++
A++2++0.285714
B++1++0.142857
C++1++0.142857
D++2++0.285714
E++1++0.142857|141858|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Try:

<pre><code>g = pd.DataFrame(['A','B','A','C','D','D','E'])

# Group by the contents of column 0 
gg = g.groupby(0) 

# Create a DataFrame with the counts of each letter
histo = gg.apply(lambda x: x.count())

# Add a new column that is the count / total number of elements 
histo[1] = histo.astype(np.float)/len(g) 

print histo
</code></pre>

Output:

<pre><code> 0 1
0 
A 2 0.285714
B 1 0.142857
C 1 0.142857
D 2 0.285714
E 1 0.142857
</code></pre>

blocks|key|141908|text|有一次，我看到一种用于计算S.O.加权平均值的嵌套函数技术，更改该技术可以解决您的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|141909|def+group_weight(overall_size):
++++def+inner(group):
++++++++return+len(group)/float(overall_size)
++++inner.__name__+=+'weight'
++++return+inner

d+=+{"my_label":+pd.Series(['A','B','A','C','D','D','E'])}
df+=+pd.DataFrame(d)
print+df.groupby('my_label').apply(group_weight(len(df)))



my_label
A++++0.285714
B++++0.142857
C++++0.142857
D++++0.285714
E++++0.142857
dtype:+float64|code-block|syntax|javascript|141910|下面是如何计算组内加权平均值的方法|141911|def+wavg(val_col_name,wt_col_name):
++++def+inner(group):
++++++++return+(group[val_col_name]+*+group[wt_col_name]).sum()+/+group[wt_col_name].sum()
++++inner.__name__+=+'wgt_avg'
++++return+inner



d+=+{"P":+pd.Series(['A','B','A','C','D','D','E'])
+++++,"Q":+pd.Series([1,2,3,4,5,6,7])
++++,"R":+pd.Series([0.1,0.2,0.3,0.4,0.5,0.6,0.7])
+++++}

df+=+pd.DataFrame(d)
print+df.groupby('P').apply(wavg('Q','R'))

P
A++++2.500000
B++++2.000000
C++++4.000000
D++++5.545455
E++++7.000000
dtype:+float64|141912|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I saw a nested function technique for computing a weighted average on S.O. one time, altering that technique can solve your issue.

<pre><code>def group_weight(overall_size):
 def inner(group):
 return len(group)/float(overall_size)
 inner.__name__ = 'weight'
 return inner

d = {"my_label": pd.Series(['A','B','A','C','D','D','E'])}
df = pd.DataFrame(d)
print df.groupby('my_label').apply(group_weight(len(df)))



my_label
A 0.285714
B 0.142857
C 0.142857
D 0.285714
E 0.142857
dtype: float64
</code></pre>

Here is how to do a weighted average within groups

<pre><code>def wavg(val_col_name,wt_col_name):
 def inner(group):
 return (group[val_col_name] * group[wt_col_name]).sum() / group[wt_col_name].sum()
 inner.__name__ = 'wgt_avg'
 return inner



d = {"P": pd.Series(['A','B','A','C','D','D','E'])
 ,"Q": pd.Series([1,2,3,4,5,6,7])
 ,"R": pd.Series([0.1,0.2,0.3,0.4,0.5,0.6,0.7])
 }

df = pd.DataFrame(d)
print df.groupby('P').apply(wavg('Q','R'))

P
A 2.500000
B 2.000000
C 4.000000
D 5.545455
E 7.000000
dtype: float64
</code></pre>

blocks|key|359830|text|除了As+of+Pandas+version+0.22，还有一种替代apply的方法：pipe，它比使用apply快得多(您还可以查看this+question了解这两种功能之间的更多差异)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|359831|对于您的示例：|359832|df+=+pd.DataFrame({"my_label":+['A','B','A','C','D','D','E']})

++my_label
0++++++++A
1++++++++B
2++++++++A
3++++++++C
4++++++++D
5++++++++D
6++++++++E|code-block|syntax|javascript|359833|apply版本|359834|df.groupby('my_label').apply(lambda+grp:+grp.count()+/+df.shape[0])|359835|给出|359836|++++++++++my_label
my_label++++++++++
A+++++++++0.285714
B+++++++++0.142857
C+++++++++0.142857
D+++++++++0.285714
E+++++++++0.142857|359837|和pipe版本|359838|df.groupby('my_label').pipe(lambda+grp:+grp.size()+/+grp.size().sum())|359839|收益率|359840|my_label
A++++0.285714
B++++0.142857
C++++0.142857
D++++0.285714
E++++0.142857|359841|因此，值是相同的，但是，计时差别很大(至少对于这个小数据帧是如此)：|359842|%25timeit+df.groupby('my_label').apply(lambda+grp:+grp.count()+/+df.shape[0])
100+loops,+best+of+3:+5.52+ms+per+loop|359843|和|359844|%25timeit+df.groupby('my_label').pipe(lambda+grp:+grp.size()+/+grp.size().sum())
1000+loops,+best+of+3:+843+µs+per+loop|359845|然后，将其封装到一个函数中也很简单：|359846|def+get_perc(grp_obj):
++++gr_size+=+grp_obj.size()
++++return+gr_size+/+gr_size.sum()|359847|现在您可以调用|359848|df.groupby('my_label').pipe(get_perc)|359849|让位|359850|my_label
A++++0.285714
B++++0.142857
C++++0.142857
D++++0.285714
E++++0.142857|359851|但是，对于这种特殊情况，您甚至不需要groupby，但可以像这样使用value_counts：|359852|df['my_label'].value_counts(sort=False)+/+df.shape[0]|359853|359854|A++++0.285714
C++++0.142857
B++++0.142857
E++++0.142857
D++++0.285714
Name:+my_label,+dtype:+float64|359855|对于这个小数据帧，它是相当快的|359856|%25timeit+df['my_label'].value_counts(sort=False)+/+df.shape[0]
1000+loops,+best+of+3:+770+µs+per+loop|359857|正如@anmol所指出的，最后一条语句也可以简化为|359858|df['my_label'].value_counts(sort=False,+normalize=True)|359859|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/version/0.22.0/whatsnew.html#groupby-objects-now-have-a-pipe-method|1|https://pandas.pydata.org/pandas-docs/version/0.22.0/groupby.html#groupby-pipe|2|https://stackoverflow.com/q/47226407/1534017^0|Y|5|17|4|1G|5|2|P|0|17|4|1|1V|D|2|0|0|0|0|5|0|0|0|0|1|4|0|0|0|0|0|0|0|0|0|0|0|0|0|0|I|7|Y|C|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|2D|8|@$9|2E|A|2F|B|C]|$9|2G|A|2H|B|C]|$9|2I|A|2J|B|C]]|D|@$9|2K|A|2L|1|2M]|$9|2N|A|2O|1|2P]|$9|2Q|A|2R|1|2S]]|E|$]]|$1|F|3|G|5|6|7|2T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|2U|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|2V|8|@$9|2W|A|2X|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|2Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|2Z|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|30|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|31|8|@$9|32|A|33|B|C]]|D|@]|E|$]]|$1|W|3|X|5|J|7|34|8|@]|D|@]|E|$K|L]]|$1|Y|3|Z|5|6|7|35|8|@]|D|@]|E|$]]|$1|10|3|11|5|J|7|36|8|@]|D|@]|E|$K|L]]|$1|12|3|13|5|6|7|37|8|@]|D|@]|E|$]]|$1|14|3|15|5|J|7|38|8|@]|D|@]|E|$K|L]]|$1|16|3|17|5|6|7|39|8|@]|D|@]|E|$]]|$1|18|3|19|5|J|7|3A|8|@]|D|@]|E|$K|L]]|$1|1A|3|1B|5|6|7|3B|8|@]|D|@]|E|$]]|$1|1C|3|1D|5|J|7|3C|8|@]|D|@]|E|$K|L]]|$1|1E|3|1F|5|6|7|3D|8|@]|D|@]|E|$]]|$1|1G|3|1H|5|J|7|3E|8|@]|D|@]|E|$K|L]]|$1|1I|3|1J|5|6|7|3F|8|@]|D|@]|E|$]]|$1|1K|3|1L|5|J|7|3G|8|@]|D|@]|E|$K|L]]|$1|1M|3|1N|5|6|7|3H|8|@$9|3I|A|3J|B|C]|$9|3K|A|3L|B|C]]|D|@]|E|$]]|$1|1O|3|1P|5|J|7|3M|8|@]|D|@]|E|$K|L]]|$1|1Q|3|1J|5|6|7|3N|8|@]|D|@]|E|$]]|$1|1R|3|1S|5|J|7|3O|8|@]|D|@]|E|$K|L]]|$1|1T|3|1U|5|6|7|3P|8|@]|D|@]|E|$]]|$1|1V|3|1W|5|J|7|3Q|8|@]|D|@]|E|$K|L]]|$1|1X|3|1Y|5|6|7|3R|8|@]|D|@]|E|$]]|$1|1Z|3|20|5|J|7|3S|8|@]|D|@]|E|$K|L]]|$1|21|3|-4|5|6|7|3T|8|@]|D|@]|E|$]]]|22|$23|$5|24|25|26|E|$27|28]]|29|$5|24|25|26|E|$27|2A]]|2B|$5|24|25|26|E|$27|2C]]]]

<a href="https://pandas.pydata.org/pandas-docs/version/0.22.0/whatsnew.html#groupby-objects-now-have-a-pipe-method" rel="noreferrer">As of Pandas version 0.22</a>, there exists also an alternative to <code>apply</code>: <a href="https://pandas.pydata.org/pandas-docs/version/0.22.0/groupby.html#groupby-pipe" rel="noreferrer"><code>pipe</code></a>, which can be considerably faster than using <code>apply</code> (you can also check <a href="https://stackoverflow.com/q/47226407/1534017">this question</a> for more differences between the two functionalities).

For your example:

<pre><code>df = pd.DataFrame({"my_label": ['A','B','A','C','D','D','E']})

 my_label
0 A
1 B
2 A
3 C
4 D
5 D
6 E
</code></pre>

The <code>apply</code> version

<pre><code>df.groupby('my_label').apply(lambda grp: grp.count() / df.shape[0])
</code></pre>

gives

<pre><code> my_label
my_label 
A 0.285714
B 0.142857
C 0.142857
D 0.285714
E 0.142857
</code></pre>

and the <code>pipe</code> version

<pre><code>df.groupby('my_label').pipe(lambda grp: grp.size() / grp.size().sum())
</code></pre>

yields

<pre><code>my_label
A 0.285714
B 0.142857
C 0.142857
D 0.285714
E 0.142857
</code></pre>

So the values are identical, however, the timings differ quite a lot (at least for this small dataframe):

<pre><code>%timeit df.groupby('my_label').apply(lambda grp: grp.count() / df.shape[0])
100 loops, best of 3: 5.52 ms per loop
</code></pre>

and

<pre><code>%timeit df.groupby('my_label').pipe(lambda grp: grp.size() / grp.size().sum())
1000 loops, best of 3: 843 µs per loop
</code></pre>

Wrapping it into a function is then also straightforward:

<pre><code>def get_perc(grp_obj):
 gr_size = grp_obj.size()
 return gr_size / gr_size.sum()
</code></pre>

Now you can call

<pre><code>df.groupby('my_label').pipe(get_perc)
</code></pre>

yielding

<pre><code>my_label
A 0.285714
B 0.142857
C 0.142857
D 0.285714
E 0.142857
</code></pre>

However, for this particular case, you do not even need a <code>groupby</code>, but you can just use <code>value_counts</code> like this:

<pre><code>df['my_label'].value_counts(sort=False) / df.shape[0]
</code></pre>

yielding

<pre><code>A 0.285714
C 0.142857
B 0.142857
E 0.142857
D 0.285714
Name: my_label, dtype: float64
</code></pre>

For this small dataframe it is quite fast

<pre><code>%timeit df['my_label'].value_counts(sort=False) / df.shape[0]
1000 loops, best of 3: 770 µs per loop
</code></pre>

As pointed out by @anmol, the last statement can also be simplified to

<pre><code>df['my_label'].value_counts(sort=False, normalize=True)
</code></pre>

blocks|key|359896|text|关于“size”的问题，size不是数据帧上的函数，而是一个属性。因此，不是使用size()，而是使用普通大小|type|unstyled|depth|inlineStyleRanges|entityRanges|data|359897|除此之外，像这样的方法应该是可行的。|359898|def+doCalculation(df):
++++groupCount+=+df.size
++++groupSum+=+df['my_labels'].notnull().sum()
++++
++++return+groupCount+/+groupSum

dataFrame.groupby('my_labels').apply(doCalculation)|code-block|syntax|javascript|359899|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Regarding the issue with 'size', size is not a function on a dataframe, it is rather a property. So instead of using size(), plain size should work
Apart from that, a method like this should work
<pre><code>def doCalculation(df):
 groupCount = df.size
 groupSum = df['my_labels'].notnull().sum()
 
 return groupCount / groupSum

dataFrame.groupby('my_labels').apply(doCalculation)
</code></pre>

I have a pandas dataframe with a column called <code>my_labels</code> which contains strings: <code>'A', 'B', 'C', 'D', 'E'</code>. I would like to count the number of occurances of each of these strings then divide the number of counts by the sum of all the counts. I'm trying to do this in Pandas like this:

<pre><code>func = lambda x: x.size() / x.sum()
data = frame.groupby('my_labels').apply(func)
</code></pre>

This code throws an error, 'DataFrame object has no attribute 'size'. How can I apply a function to calculate this in Pandas?

Apply function to pandas groupby

我有一个pandas数据帧，它有一个名为my_labels的列，其中包含字符串：'A', 'B', 'C', 'D', 'E'。我想计算每个字符串出现的次数，然后将计数的数量除以所有计数的总和。我试着在Pandas中这样做：func = lambda x: x.size() / x.sum()data = frame....

问Apply function to pandas groupby
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