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pandas的Series类型与基本操作详解

作者:kingov

这篇文章主要介绍了pandas的Series类型与基本操作详解,文中通过示例代码介绍的非常详细,对大家的学习或者工作具有一定的参考学习价值,需要的朋友们下面随着小编来一起学习学习吧

1 Series

线性的数据结构, series是一个一维数组

Pandas 会默然用0到n-1来作为series的index, 但也可以自己指定index( 可以把index理解为dict里面的key )

1.1创造一个serise数据

import pandas as pd
import numpy as np
​s = pd.Series([9, 'zheng', 'beijing', 128])
​print(s)

打印

0 9
1 zheng
2 beijing
3 128
dtype: object

访问其中某个数据

print(s[1:2])
​
# 打印
1 zheng
dtype: object

Series类型的基本操作:

Series类型包括index和values两部分

In [14]: a = pd.Series({'a':1,'b':5})

In [15]: a.index
Out[15]: Index(['a', 'b'], dtype='object')

In [16]: a.values #返回一个多维数组numpy对象
Out[16]: array([1, 5], dtype=int64)

Series类型的操作类似ndarray类型

#自动索引和自定义索引并存,但不能混用
In [17]: a[0] #自动索引
Out[17]: 1
#自定义索引
In [18]: a['a']
Out[18]: 1
#不能混用
In [20]: a[['a',1]]
Out[20]:
a 1.0
1 NaN
dtype: float64

Series类型的操作类似Python字典类型

#通过自定义索引访问
#对索引保留字in操作,值不可以
In [21]: 'a' in a
Out[21]: True

In [22]: 1 in a
Out[22]: False

Series类型在运算中会自动对齐不同索引的数据

In [29]: a = pd.Series([1,3,5],index = ['a','b','c'])

In [30]: b = pd.Series([2,4,5,6],index = ['c,','d','e','b'])

In [31]: a+b
Out[31]:
a  NaN
b  9.0
c  NaN
c, NaN
d  NaN
e  NaN
dtype: float64

Series对象可以随时修改并即刻生效

In [32]: a.index = ['c','d','e']

In [33]: a
Out[33]:
c 1
d 3
e 5
dtype: int64

In [34]: a+b
Out[34]:
b  NaN
c  NaN
c,  NaN
d  7.0
e  10.0
dtype: float64

1.2 指定index

import pandas as pd
import numpy as np
​s = pd.Series([9, 'zheng', 'beijing', 128, 'usa', 990], index=[1,2,3,'e','f','g'])​
print(s)

打印

1 9
2 zheng
3 beijing
e 128
f usa
g 990
dtype: object

根据索引找出值

print(s['f']) # usa

1.3 用dictionary构造一个series

import pandas as pd
import numpy as np
s = {"ton": 20, "mary": 18, "jack": 19, "car": None}
sa = pd.Series(s, name="age")​
print(sa)

打印

car NaN
jack 19.0
mary 18.0
ton 20.0
Name: age, dtype: float64

检测类型

print(type(sa)) # <class 'pandas.core.series.Series'>

1.4 用numpy ndarray构造一个Series

生成一个随机数

import pandas as pd
import numpy as np
​
num_abc = pd.Series(np.random.randn(5), index=list('abcde'))
num = pd.Series(np.random.randn(5))
​
print(num)
print(num_abc)
​
# 打印
0   -0.102860
1   -1.138242
2    1.408063
3   -0.893559
4    1.378845
dtype: float64
a   -0.658398
b    1.568236
c    0.535451
d    0.103117
e   -1.556231
dtype: float64

1.5 选择数据

import pandas as pd
import numpy as np
​
s = pd.Series([9, 'zheng', 'beijing', 128, 'usa', 990], index=[1,2,3,'e','f','g'])
​
print(s[1:3])  # 选择第1到3个, 包左不包右 zheng beijing
print(s[[1,3]])  # 选择第1个和第3个, zheng 128
print(s[:-1]) # 选择第1个到倒数第1个, 9 zheng beijing 128 usa

1.6 操作数据

import pandas as pd
import numpy as np​
s = pd.Series([9, 'zheng', 'beijing', 128, 'usa', 990], index=[1,2,3,'e','f','g'])

sum = s[1:3] + s[1:3]
sum1 = s[1:4] + s[1:4]
sum2 = s[1:3] + s[1:4]
sum3 = s[:3] + s[1:]​
print(sum)
print(sum1)
print(sum2)
print(sum3)

打印

2        zhengzheng
3    beijingbeijing
dtype: object
2        zhengzheng
3    beijingbeijing
e               256
dtype: object
2        zhengzheng
3    beijingbeijing
e               NaN
dtype: object
1               NaN
2        zhengzheng
3    beijingbeijing
e               NaN
f               NaN
g               NaN
dtype: object

1.7 查找

是否存在

USA in s # true

范围查找

import pandas as pd
import numpy as np
 
s = {"ton": 20, "mary": 18, "jack": 19, "jim": 22, "lj": 24, "car": None}
 
sa = pd.Series(s, name="age")
 
print(sa[sa>19])


中位数

import pandas as pd
import numpy as np
 
s = {"ton": 20, "mary": 18, "jack": 19, "jim": 22, "lj": 24, "car": None}
 
sa = pd.Series(s, name="age")
 
print(sa.median()) # 20

判断是否大于中位数

import pandas as pd
import numpy as np
 
s = {"ton": 20, "mary": 18, "jack": 19, "jim": 22, "lj": 24, "car": None}
 
sa = pd.Series(s, name="age")
 
print(sa>sa.median())


找出大于中位数的数

import pandas as pd
import numpy as np
 
s = {"ton": 20, "mary": 18, "jack": 19, "jim": 22, "lj": 24, "car": None}
 
sa = pd.Series(s, name="age")
 
print(sa[sa > sa.median()])


中位数

import pandas as pd
import numpy as np
 
s = {"ton": 20, "mary": 18, "jack": 19, "jim": 22, "lj": 24, "car": None}
 
sa = pd.Series(s, name="age")
 
more_than_midian = sa>sa.median()
 
print(more_than_midian)
 
print('---------------------')
 
print(sa[more_than_midian])


1.8 Series赋值

import pandas as pd
import numpy as np
 
s = {"ton": 20, "mary": 18, "jack": 19, "jim": 22, "lj": 24, "car": None}
 
sa = pd.Series(s, name="age")
 
print(s)
 
print('----------------')
 
sa['ton'] = 99
 
print(sa)


1.9 满足条件的统一赋值

import pandas as pd
import numpy as np
 
s = {"ton": 20, "mary": 18, "jack": 19, "jim": 22, "lj": 24, "car": None}
 
sa = pd.Series(s, name="age")
 
print(s) # 打印原字典
 
print('---------------------') # 分割线
 
sa[sa>19] = 88 # 将所有大于19的同一改为88
 
print(sa) # 打印更改之后的数据
 
print('---------------------') # 分割线
 
print(sa / 2) # 将所有数据除以2

到此这篇关于pandas的Series类型与基本操作详解的文章就介绍到这了,更多相关pandas Series基本操作内容请搜索脚本之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持脚本之家!

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