处理丢失数据

有两种丢失数据:

  • None
  • np.nan(NaN)
import numpy as np

type(None)
NoneType
type(np.nan)
float

1. None

None是Python自带的,其类型为python object。因此,None不能参与到任何计算中。

object类型的运算要比int类型的运算慢得多

计算不同数据类型求和时间

%timeit np.arange(1e5,dtype=xxx).sum()

1E7
10000000.0
%timeit np.arange(1E6, dtype= int).sum()
4.85 ms ± 1.08 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit np.arange(1E6, dtype = float).sum()
7.33 ms ± 587 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit np.arange(1E6,dtype = object).sum()
118 ms ± 8.94 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

2. np.nan(NaN)

np.nan是浮点类型,能参与到计算中。但计算的结果总是NaN。

但可以使用np.nan*()函数来计算nan,此时视nan为0。

nd = np.array([10,20,30,np.nan,None])
#None 不能够参加到运算当中
nd.sum()
---------------------------------------------------------------------------

TypeError                                 Traceback (most recent call last)

<ipython-input-13-218f693fc17f> in <module>()
1 nd = np.array([10,20,30,np.nan,None])
----> 2 nd.sum() C:\anaconda\lib\site-packages\numpy\core\_methods.py in _sum(a, axis, dtype, out, keepdims)
30
31 def _sum(a, axis=None, dtype=None, out=None, keepdims=False):
---> 32 return umr_sum(a, axis, dtype, out, keepdims)
33
34 def _prod(a, axis=None, dtype=None, out=None, keepdims=False): TypeError: unsupported operand type(s) for +: 'float' and 'NoneType'
nd = np.array([10,20,30,np.nan])
nd
array([10., 20., 30., nan])
nd.sum()
nan
np.mean(nd)
nan
np.nanmean(nd)
20.0
np.nansum(nd)
60.0
np.nan
nan

3. pandas中的None与NaN

1) pandas中None与np.nan都视作np.nan

创建DataFrame

import pandas as pd
from pandas import Series,DataFrame
df = DataFrame([10,20,57,None,np.nan], index = list('abcde'), columns = ["Python"])
df

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Python
a 10.0
b 20.0
c 57.0
d NaN
e NaN
df.sum()
Python    87.0
dtype: float64
df = DataFrame([[10,20,57,None,np.nan],
[22,33,56,16,None],
[np.nan,1,2,3,4]], index = list("abc"), columns = ["Python","Java","物理","数学","H5"])
df

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Python Java 物理 数学 H5
a 10.0 20 57 NaN NaN
b 22.0 33 56 16.0 NaN
c NaN 1 2 3.0 4.0
df.sum(axis = 0)
Python     32.0
Java 54.0
物理 115.0
数学 19.0
H5 4.0
dtype: float64

使用DataFrame行索引与列索引修改DataFrame数据

df["Python"]["c"] = 12
C:\anaconda\lib\site-packages\ipykernel_launcher.py:1: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
"""Entry point for launching an IPython kernel.
df

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Python Java 物理 数学 H5
a 10.0 20 57 NaN NaN
b 22.0 33 56 16.0 NaN
c 12.0 1 2 3.0 4.0

2) pandas中None与np.nan的操作

df = DataFrame([[10,20,57,None,np.nan],
[22,33,56,16,None],
[np.nan,1,2,3,4]], index = list("abc"), columns = ["Python","Java","物理","数学","H5"])
df

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Python Java 物理 数学 H5
a 10.0 20 57 NaN NaN
b 22.0 33 56 16.0 NaN
c NaN 1 2 3.0 4.0
#下面讲的是一个重点!!!!
  • isnull()
  • notnull()
  • dropna(): 过滤丢失数据
  • fillna(): 填充丢失数据
df

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Python Java 物理 数学 H5
a 10.0 20 57 NaN NaN
b 22.0 33 56 16.0 NaN
c NaN 1 2 3.0 4.0
#DataFrame 的isnull这个函数返回值就是一个DataFrame
is_null = df.isnull()
is_null
#

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Python Java 物理 数学 H5
a False False False True True
b False False False False True
c True False False False False
#需求:查看哪一行有空值,举数据分析的例子的时候吗,会用这个方法
is_null = is_null.any(axis = 1)
is_null
a    True
b True
c True
dtype: bool
df2 = DataFrame([[10,20,57,90,28],[22,35,46,78,67],[21,34,23,77,66]],
index = list("efg"),columns = ["Python","Java","物理","数学","H5"] )
#没空值的数据
df2

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Python Java 物理 数学 H5
e 10 20 57 90 28
f 22 35 46 78 67
g 21 34 23 77 66
df3 = df.add(df2, fill_value = 0)
df3

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 NaN NaN
b 22.0 33.0 56.0 16.0 NaN
c NaN 1.0 2.0 3.0 4.0
e 10.0 20.0 57.0 90.0 28.0
f 22.0 35.0 46.0 78.0 67.0
g 21.0 34.0 23.0 77.0 66.0
df3_isnull = df3.isnull()
df3_isnull = df3_isnull.any(axis = 1)
df3_isnull
a     True
b True
c True
e False
f False
g False
dtype: bool
df3[df3_isnull]
#过滤问题 过滤的是没有空值的,留下来的是带空值的!!!

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 NaN NaN
b 22.0 33.0 56.0 16.0 NaN
c NaN 1.0 2.0 3.0 4.0
df

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Python Java 物理 数学 H5
a 10.0 20 57 NaN NaN
b 22.0 33 56 16.0 NaN
c NaN 1 2 3.0 4.0
df[is_null]

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Python Java 物理 数学 H5
a 10.0 20 57 NaN NaN
b 22.0 33 56 16.0 NaN
c NaN 1 2 3.0 4.0

(1)判断函数

  • isnull()
  • notnull()
df3

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 NaN NaN
b 22.0 33.0 56.0 16.0 NaN
c NaN 1.0 2.0 3.0 4.0
e 10.0 20.0 57.0 90.0 28.0
f 22.0 35.0 46.0 78.0 67.0
g 21.0 34.0 23.0 77.0 66.0
df3_notnull = df3.notnull().all(axis = 1)
df3_notnull
a    False
b False
c False
e True
f True
g True
dtype: bool
df3[df3_notnull]
#过滤的是空值,留下来的是没有空值的情况

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Python Java 物理 数学 H5
e 10.0 20.0 57.0 90.0 28.0
f 22.0 35.0 46.0 78.0 67.0
g 21.0 34.0 23.0 77.0 66.0
#还可以通过条件来进行过滤
df3

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 NaN NaN
b 22.0 33.0 56.0 16.0 NaN
c NaN 1.0 2.0 3.0 4.0
e 10.0 20.0 57.0 90.0 28.0
f 22.0 35.0 46.0 78.0 67.0
g 21.0 34.0 23.0 77.0 66.0
cond = (df3 >= 10).all(axis= 1)
cond
a    False
b False
c False
e True
f True
g True
dtype: bool
df3[cond]

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Python Java 物理 数学 H5
e 10.0 20.0 57.0 90.0 28.0
f 22.0 35.0 46.0 78.0 67.0
g 21.0 34.0 23.0 77.0 66.0

(2) 过滤函数

  • dropna()
df3

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 NaN NaN
b 22.0 33.0 56.0 16.0 NaN
c NaN 1.0 2.0 3.0 4.0
e 10.0 20.0 57.0 90.0 28.0
f 22.0 35.0 46.0 78.0 67.0
g 21.0 34.0 23.0 77.0 66.0
df3.dropna()

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Python Java 物理 数学 H5
e 10.0 20.0 57.0 90.0 28.0
f 22.0 35.0 46.0 78.0 67.0
g 21.0 34.0 23.0 77.0 66.0
df3["H5"] = None
df3
#pandas 自身的bug 但是数据还是nan

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 NaN None
b 22.0 33.0 56.0 16.0 None
c NaN 1.0 2.0 3.0 None
e 10.0 20.0 57.0 90.0 None
f 22.0 35.0 46.0 78.0 None
g 21.0 34.0 23.0 77.0 None
df3.dropna(axis = 1,how = "all")

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Python Java 物理 数学
a 10.0 20.0 57.0 NaN
b 22.0 33.0 56.0 16.0
c NaN 1.0 2.0 3.0
e 10.0 20.0 57.0 90.0
f 22.0 35.0 46.0 78.0
g 21.0 34.0 23.0 77.0

可以选择过滤的是行还是列(默认为行)

也可以选择过滤的方式 how = 'all'

(3) 填充函数 Series/DataFrame

  • fillna()
df3

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 NaN None
b 22.0 33.0 56.0 16.0 None
c NaN 1.0 2.0 3.0 None
e 10.0 20.0 57.0 90.0 None
f 22.0 35.0 46.0 78.0 None
g 21.0 34.0 23.0 77.0 None
df3.fillna(-1)

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 -1.0 -1
b 22.0 33.0 56.0 16.0 -1
c -1.0 1.0 2.0 3.0 -1
e 10.0 20.0 57.0 90.0 -1
f 22.0 35.0 46.0 78.0 -1
g 21.0 34.0 23.0 77.0 -1

可以选择前向填充还是后向填充

df3

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 NaN None
b 22.0 33.0 56.0 16.0 None
c NaN 1.0 2.0 3.0 None
e 10.0 20.0 57.0 90.0 None
f 22.0 35.0 46.0 78.0 None
g 21.0 34.0 23.0 77.0 None
df3.fillna(method = "bfill")

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 16.0 None
b 22.0 33.0 56.0 16.0 None
c 10.0 1.0 2.0 3.0 None
e 10.0 20.0 57.0 90.0 None
f 22.0 35.0 46.0 78.0 None
g 21.0 34.0 23.0 77.0 None
df3.fillna(method = "ffill")

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 NaN None
b 22.0 33.0 56.0 16.0 None
c 22.0 1.0 2.0 3.0 None
e 10.0 20.0 57.0 90.0 None
f 22.0 35.0 46.0 78.0 None
g 21.0 34.0 23.0 77.0 None
#f  forward  向前
df3.fillna(method='ffill', axis = 1)

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Python Java 物理 数学 H5
a 10.0 20.0 57.0 57.0 57.0
b 22.0 33.0 56.0 16.0 16.0
c NaN 1.0 2.0 3.0 3.0
e 10.0 20.0 57.0 90.0 90.0
f 22.0 35.0 46.0 78.0 78.0
g 21.0 34.0 23.0 77.0 77.0
df3.fillna(method = "bfill",axis = 1)

对于DataFrame来说,还要选择填充的轴axis。记住,对于DataFrame来说:

  • axis=0:index/行
  • axis=1:columns/列

============================================

练习7:

  1. 简述None与NaN的区别

  2. 假设张三李四参加模拟考试,但张三因为突然想明白人生放弃了英语考试,因此记为None,请据此创建一个DataFrame,命名为ddd3

  3. 老师决定根据用数学的分数填充张三的英语成绩,如何实现?

    用李四的英语成绩填充张三的英语成绩?

============================================

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