pandsa_series02

7. 如何获得数值series的四分位值

   \# 设置随机数种子
   state = np.random.RandomState(100)
   \# 从均值为5标准差为25的正态分布随机抽取5个点构成series
   ser = pd.Series(state.normal(10, 5, 25))
   \# 求ser的四分位数
   
       np.percentile(ser, q=[0, 25, 50, 75, 100])
       
       #> array([ 1.25117263,  7.70986507, 10.92259345, 13.36360403, 18.0949083 ])

   8. 如何获得series中单一项的频率计数

      #从0~7随机抽取30个列表值，组成series
      ser = pd.Series(np.take(list('abcdefgh'), np.random.randint(8, size=30)))
      # 对该series进行计数
      
          ser.value_counts()
          
          #>    d    8
              g    6
              b    6
              a    5
              e    2
              h    2
              f    1
              dtype: int64

      9. 如何保留series中前两个频次最多的项，其他项替换为‘other’

      10. np.random.RandomState(100)
          \# 从1~4均匀采样12个点组成series
            ser = pd.Series(np.random.randint(1, 5, [12]))
              # 除前两行索引对应的值不变，后几行索引对应的值为Other
              ser[~ser.isin(ser.value_counts().index[:2])] = 'Other'
              ser
              
            #>    0     Other
                 1         4
                 2         2
                 3         2
                 4         4
                 5     Other
                 6     Other
                 7     Other
                 8         4
                 9         4
                 10        4
                 11        2
                 dtype: object

          10. 如何对数值series分成10个相同数目的组
              换个角度理解，对数值series离散化成10个类别（categorical）值

          ser = pd.Series(np.random.random(20))
          
          \# 离散化10个类别值,只显示前5行的数据
          
             pd.qcut(ser, q=[0, .10, .20, .3, .4, .5, .6, .7, .8, .9, 1], 
                     labels=['1st', '2nd', '3rd', '4th', '5th', '6th', '7th', '8th', '9th', '10th']).head()
             
             #>
             0    3rd
             1    1st
             2    6th
             3    6th
             4    9th
             dtype: category
             Categories (10, object): [1st < 2nd < 3rd < 4th ... 7th < 8th < 9th < 10th]

          11. 如何使numpy数组转化为给定形状的dataframe

              ser = pd.Series(np.random.randint(1, 10, 35))
              \# serier类型转换numpy类型，然后重构
              
                 df = pd.DataFrame(ser.values.reshape(7,5))
                 print(df)
                 
                 #>       0  1  2  3  4
                     0  1  2  1  2  5
                     1  1  2  4  5  2
                     2  1  3  3  2  8
                     3  8  6  4  9  6
                     4  2  1  1  8  5
                     5  3  2  8  5  6
                     6  1  5  5  4  6

              12. 如何找到series的值是3的倍数的位置

                  ser = pd.Series(np.random.randint(1, 10, 7))
                  print(ser)
                  
                  # 获取值是3倍数的索引
                  
                      np.argwhere(ser % 3==0)
                      
                      #>    0    6
                          1    8
                          2    6
                          3    7
                          4    6
                          5    2
                          6    4
                          dtype: int64
                          
                      #>    array([[0],
                                 [2],
                                 [4]])