大熊猫任意分布的频率均值计算

Question

我有一个大型数据集，其值范围从1 to 25，分辨率为o.1。分布本质上是任意的，模式值为1.样本数据集可以是：

1,
1,
23.05,
19.57,
1,
1.56,
1,
23.53,
19.74,
7.07,
1,
22.85,
1,
1,
7.78,
16.89,
12.75,
15.32,
7.7,
14.26,
15.41,
1,
16.34,
8.57,
15,
14.97,
1.18,
14.15,
1.94,
14.61,
1,
15.49,
1,
9.18,
1.71,
1,
10.4,

如何评估不同范围（0-0.5,0.5-1等）的计数，并在pandas中找出它们的频率均值，Python。

预期产量可以

值的范围（of）occurrence（n）f * n

 1  
 2.2      1-2   2   3
 2.8      2-3   3   7.5
 3.7      3-4   2   7
 5.5      4-5   1   4.5
 5.8      5-6   3   16.5
 4.3            
 2.7 sum-  11        38.5
 3.5            
 1.8        frequency mean  3.5
 5.9            

Answer 1

您需要cut进行分级，然后将CategoricalIndex转换为IntervalIndex以获取mid值，将mul，sum和last div标量转换为多列：

df = pd.DataFrame({'col':[1,2.2,2.8,3.7,5.5,5.8,4.3,2.7,3.5,1.8,5.9]})
print (df)
    col
0   1.0
1   2.2
2   2.8
3   3.7
4   5.5
5   5.8
6   4.3
7   2.7
8   3.5
9   1.8
10  5.9

---

binned = pd.cut(df['col'], np.arange(1, 7), include_lowest=True)
df1 = df.groupby(binned).size().reset_index(name='val')
df1['mid'] = pd.IntervalIndex(df1['col']).mid
df1['mul'] = df1['val'].mul(df1['mid'])
print (df1)
            col  val     mid     mul
0  (0.999, 2.0]    2  1.4995   2.999
1    (2.0, 3.0]    3  2.5000   7.500
2    (3.0, 4.0]    2  3.5000   7.000
3    (4.0, 5.0]    1  4.5000   4.500
4    (5.0, 6.0]    3  5.5000  16.500

a = df1.sum()
print (a)
val    11.0000
mid    17.4995
mul    38.4990
dtype: float64

b = a['mul'] / a['val']
print (b)
3.49990909091