目录
- 1 Series
- 1.1创造一个serise数据
- 1.2 指定index
- 1.3 用dictionary构造一个series
- 1.4 用numpy ndarray构造一个Series
- 1.5 选择数据
- 1.6 操作数据
- 1.7 查找
- 1.8 Series赋值
- 1.9 满足条件的统一赋值
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
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
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
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