Pandas：根据特定列的值计数选择行
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Pandas: Selecting rows based on value counts of a particular column
提问by bigO6377
Whats the simplest way of selecting all rows from a panda dataframe, who's sym occurs exactly twice in the entire table? For example, in the table below, I would like to select all rows with sym in ['b','e'], since the value_counts for these symbols equal 2.
从Pandas数据框中选择所有行的最简单方法是什么，谁的 sym 在整个表中恰好出现两次？例如，在下表中，我想选择 ['b','e'] 中带有 sym 的所有行，因为这些符号的 value_counts 等于 2。
df=pd.DataFrame({'sym':['a', 'b', 'b', 'c', 'd','d','d','e','e'],'price':np.random.randn(9)})
price sym
0 -0.0129 a
1 -1.2940 b
2 1.8423 b
3 -0.7160 c
4 -2.3216 d
5 -0.0120 d
6 -0.5914 d
7 0.6280 e
8 0.5361 e
df.sym.value_counts()
Out[237]:
d 3
e 2
b 2
c 1
a 1
回答by jezrael
I think you can use groupbyby column symand filtervalues with length == 2:
我认为您可以groupby按列sym和filter值使用length == 2：
print df.groupby("sym").filter(lambda x: len(x) == 2)
price sym
1 0.400157 b
2 0.978738 b
7 -0.151357 e
8 -0.103219 e
Second solution use isinwith boolean indexing:
s = df.sym.value_counts()
print s[s == 2].index
Index([u'e', u'b'], dtype='object')
print df[df.sym.isin(s[s == 2].index)]
price sym
1 0.400157 b
2 0.978738 b
7 -0.151357 e
8 -0.103219 e
And fastest solution with transformand boolean indexing:
并用最快的解决方案transform和boolean indexing：
print (df[df.groupby("sym")["sym"].transform('size') == 2])
price sym
1 -1.2940 b
2 1.8423 b
7 0.6280 e
8 0.5361 e
回答by Tim Cui
You can use map, which should be faster than using groupbyand transform:
您可以使用map，这应该比使用groupbyand更快transform：
df[df['sym'].map(df['sym'].value_counts()) == 2]
e.g.
例如
%%timeit
df[df['sym'].map(df['sym'].value_counts()) == 2]
Out[1]:
1.83 ms ± 23.7 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
df[df.groupby("sym")["sym"].transform('size') == 2]
Out[2]:
2.08 ms ± 41.3 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
