数据准备：
In [1]:import numpy as np
In [2]:t = np.arange(24).reshape(4,6) # 创建数组
In [3]:t
Out[3]:
array([[ 0, 1, 2, 3, 4, 5],
[ 6, 7, 8, 9, 10, 11],
[12, 13, 14, 15, 16, 17],
[18, 19, 20, 21, 22, 23]])
In [4]:t = t.astype(float) # 修改数组的数据类型为float
In [5]: t
Out[5]:
array([[ 0., 1., 2., 3., 4., 5.],
[ 6., 7., 8., 9., 10., 11.],
[12., 13., 14., 15., 16., 17.],
[18., 19., 20., 21., 22., 23.]])
In [6]:t[3,3:] = np.nan # 赋值nan
In [7]:t
Out[7]:
array([[ 0., 1., 2., 3., 4., 5.],
[ 6., 7., 8., 9., 10., 11.],
[12., 13., 14., 15., 16., 17.],
[18., 19., 20., nan, nan, nan]])
- 求和
In [8]:np.sum(t,axis=0)
Out[8]: array([36., 40., 44., nan, nan, nan])
In [9]:np.sum(t,axis=1)
Out[9]: array([15., 51., 87., nan])
In [10]:t.sum(axis=0)
Out[10]: array([36., 40., 44., nan, nan, nan])
In [11]:t.sum(axis=1)
Out[11]: array([15., 51., 87., nan])
- 求均值
In [12]:t.mean()
Out[12]: nan
In [13]:t.mean(axis=0)
Out[13]: array([ 9., 10., 11., nan, nan, nan])
In [14]:t.mean(axis=1)
Out[14]: array([ 2.5, 8.5, 14.5, nan])
- 求中值
In [15]: t.median() # 无该种求中值类型
Traceback (most recent call last):
File "", line 1, in
t.median()
AttributeError: 'numpy.ndarray' object has no attribute 'median'
In [16]:np.median(t,axis=0)
Out[16]: array([ 9., 10., 11., nan, nan, nan])
In [17]:np.median(t,axis=1)
Out[17]: array([ 2.5, 8.5, 14.5, nan])
- 求最大值和最小值
# 求最大值
In [18]:np.max(t,axis=0) # 方案一
Out[18]: array([18., 19., 20., nan, nan, nan])
In [19]:np.max(t,axis=1)
Out[19]: array([ 5., 11., 17., nan])
**********************************************
In [20]:t.max(axis=0) # 方案二
Out[20]: array([18., 19., 20., nan, nan, nan])
In [21]:t.max(axis=1)
Out[21]: array([ 5., 11., 17., nan])
===========================================
# 求最小值
In [22]:np.min(t,axis=0)
Out[22]: array([ 0., 1., 2., nan, nan, nan])
In [23]:np.min(t,axis=1)
Out[23]: array([ 0., 6., 12., nan])
********************************************
In [24]:t.min(axis=0)
Out[24]: array([ 0., 1., 2., nan, nan, nan])
In [25]:t.min(axis=1)
Out[25]: array([ 0., 6., 12., nan])
- 求极值（最大值和最小值之差）、
In [26]:t.ptp(axis=0) # 两种方案，都可以求得结果
Out[26]: array([18., 18., 18., nan, nan, nan])
In [29]:np.ptp(t,axis=0)
Out[29]: array([18., 18., 18., nan, nan, nan])
In [27]:t.ptp(axis=1)
Out[27]: array([ 5., 5., 5., nan])
In [30]:np.ptp(t,axis=1)
Out[30]: array([ 5., 5., 5., nan])
6、标准差
In [31]:np.std(t,axis=0) # # 两种方案，都可以求得结果
Out[31]:
array([6.70820393, 6.70820393, 6.70820393, nan, nan,
nan])
In [33]:np.std(t,axis=1)
Out[33]:
array([6.70820393, 6.70820393, 6.70820393, nan, nan,
nan])
In [32]:np.std(t,axis=1)
Out[32]: array([1.70782513, 1.70782513, 1.70782513, nan])
In [34]:t.std(axis=1)
Out[34]: array([1.70782513, 1.70782513, 1.70782513, nan])