[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
data = np.array(['a','b','c','d'])
s = pd.Series(data)
print(s)
执行上面代码,结果如下:
这里没有传递任何索引,因此默认情况下,它分配了从 0 到 len(data)-1 的索引,即:0到3
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
data = np.array(['a','b','c','d'])
s = pd.Series(data,index=[100,101,102,103])
print(s)
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
data = {'a' : 0., 'b' : 1., 'c' : 2.}
s = pd.Series(data)
print(s)
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
data = {'a' : 0., 'b' : 1., 'c' : 2.}
s = pd.Series(data,index=['b','c','d','a'])
print(s)
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
s = pd.Series(5, index=[0, 1, 2, 3])
print(s)
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
g_values = np.arange(5)
g_index = np.arange(9, 4, -1)
g = pd.Series(g_values, index = g_index)
print(g)
执行上面代码,返回结果如下:
[Python] 纯文本查看 复制代码
h = pd.Series(range(10))
print(h)
执行上面代码,结果如下:
2,查询
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
data = {'a' : 0., 'b' : 1., 'c' : 2.}
s = pd.Series(data,index=['b','c','d','a'])
print(s)
print('**'*10)
print(s.values)
print('**'*10)
print(s.index)
print('**'*10)
print(s.index[2])
print('**'*10)
print(s.index.values)
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
data = {'a' : 0., 'b' : 1., 'c' : 2.}
s = pd.Series(data,index=['b','c','d','a'])
print(s)
print('**'*10)
# 查询单值
print(s[1]) # 位置索引,默认,自动生成,和自定义索引并存
print('**'*10)
print(s['a']) # 标签索引,自定义索引
print('**'*10)
print(s.d) # 方法调用
print('**'*10)
# 查询多值
print(s[[0,3]])
print('**'*10)
print(s[['a', 'd']])
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
data = {'a' : 0., 'b' : 1., 'c' : 2., 'f': 12., 'q': 3.4, 'y': 5.6}
s = pd.Series(data,index=['b','c','d','a','f','q','y','e'])
print(s)
print('**'*10)
print(s[:4]) # 位置切片,默认索引,左闭右开
print('**'*10)
print(s[:'q']) # 标签切片,自定义索引, 注意:两边都闭区间(因为使用标签索引时通常不知道标签顺序,很难确定结束前一个标签是什么)
print('**'*10)
print(s['f':])
print('**'*10)
print(s[::2])
print('**'*10)
print(s[::-1]) #步长-1,逆序
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
s = pd.Series([100,25,59,90,61],index=['ming','hua','hong','huang','bai'])
print(s)
print('**'*10)
print(s>=60)
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
s = pd.Series([100,25,59,90,61],index=['ming','hua','hong','huang','bai'])
print(s)
print('**'*10)
print(s['hua']>60)
print(s['ming']>60)
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
s = pd.Series([100,25,59,90,61],index=['ming','hua','hong','huang','bai'])
print(s)
print('**'*10)
print(s+100) # 标量运算
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
s = pd.Series([100,25,59,90,61],index=['ming','hua','hong','huang','bai'])
print(s)
print('**'*10)
print(np.median(s)) # 应用函数 , 获取中位数
print(s.median()) # 方法调用的写法
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
s = pd.Series([100,25,59,90,61],index=['ming','hua','hong','huang','bai'])
print(s)
print('**'*10)
res = 'hua' in s
print(res)
print('**'*10)
print(s.get('bai', 80))
print(s.get('hei', 80))
执行上面代码,结果如下:
3,修改
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
s = pd.Series([100,25,59,90,61],index=['ming','hua','hong','huang','bai'])
print(s)
print('**'*10)
s['ming'] = 0
print(s)
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
s = pd.Series([100,25,59,90,61],index=['ming','hua','hong','huang','bai'])
print(s)
print('**'*10)
s['ming', 'hua'] = 0
print(s)
print('**'*10)
s['ming', 'hua'] = [30, 40]
print(s)
执行上面代码,结果如下:
[Python] 纯文本查看 复制代码
import pandas as pd
import numpy as np
s = pd.Series([100,25,59,90,61],index=['ming','hua','hong','huang','bai'])
print(s)
print('**'*10)
s.index = ['xiaoming','xiaohua','xiaohong','xiaohuang','xiaobai'] # 索引修改
print(s)
print('**'*10)
s.index.values[2] = 'laoli'
print(s)
执行上面代码,结果如下:
作者: baby14 时间: 2018-6-10 08:51
多谢分享
作者: 吴琼老师 时间: 2018-7-5 16:47
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