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pandas的Series類型與基本操作詳解

更新時間：2021年03月19日 10:05:24 作者：kingov

這篇文章主要介紹了pandas的Series類型與基本操作詳解，文中通過示例代碼介紹的非常詳細，對大家的學習或者工作具有一定的參考學習價值，需要的朋友們下面隨著小編來一起學習學習吧

1 Series

線性的數(shù)據(jù)結(jié)構(gòu), series是一個一維數(shù)組

Pandas 會默然用0到n-1來作為series的index, 但也可以自己指定index( 可以把index理解為dict里面的key )

1.1創(chuàng)造一個serise數(shù)據(jù)

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

訪問其中某個數(shù)據(jù)

print(s[1:2])

# 打印
1 zheng
dtype: object

Series類型的基本操作：

Series類型包括index和values兩部分

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

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

In [16]: a.values #返回一個多維數(shù)組numpy對象
Out[16]: array([1, 5], dtype=int64)

Series類型的操作類似ndarray類型

#自動索引和自定義索引并存，但不能混用
In [17]: a[0] #自動索引
Out[17]: 1
#自定義索引
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類型的操作類似Python字典類型

#通過自定義索引訪問
#對索引保留字in操作，值不可以
In [21]: 'a' in a
Out[21]: True

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

Series類型在運算中會自動對齊不同索引的數(shù)據(jù)

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對象可以隨時修改并即刻生效

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

根據(jù)索引找出值

print(s['f']) # usa

1.3 用dictionary構(gòu)造一個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

檢測類型

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

1.4 用numpy ndarray構(gòu)造一個Series

生成一個隨機數(shù)

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 選擇數(shù)據(jù)

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])  # 選擇第1到3個, 包左不包右 zheng beijing
print(s[[1,3]])  # 選擇第1個和第3個, zheng 128
print(s[:-1]) # 選擇第1個到倒數(shù)第1個, 9 zheng beijing 128 usa

1.6 操作數(shù)據(jù)

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

范圍查找

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])

中位數(shù)

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

判斷是否大于中位數(shù)

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())

找出大于中位數(shù)的數(shù)

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()])

中位數(shù)

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賦值

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 滿足條件的統(tǒng)一賦值

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('---------------------') # 分割線
 
sa[sa>19] = 88 # 將所有大于19的同一改為88
 
print(sa) # 打印更改之后的數(shù)據(jù)
 
print('---------------------') # 分割線
 
print(sa / 2) # 將所有數(shù)據(jù)除以2