pandas的Series類型與基本操作詳解
1 Series
線性的數(shù)據(jù)結(jié)構(gòu), series是一個(gè)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(gè)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(gè)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(gè)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(yùn)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(shí)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(gè)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(gè)Series
生成一個(gè)隨機(jī)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(gè), 包左不包右 zheng beijing print(s[[1,3]]) # 選擇第1個(gè)和第3個(gè), zheng 128 print(s[:-1]) # 選擇第1個(gè)到倒數(shù)第1個(gè), 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
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