Question

我具有以下格式的各家商店的每周数据：

pd.DataFrame({'Store':['S1', 'S1', 'S1', 'S2','S2','S2','S3','S3','S3'], 'Week':[1, 2, 3,1,2,3,1,2,3],
                           'Sales' : [20,30,40,21,31,41,22,32,42],'Cust_count' : [2,4,6,3,5,7,4,6,8]})

   Store Week Sales Cust_count
0   S1   1    20    2
1   S1   2    30    4
2   S1   3    40    6
3   S2   1    21    3
4   S2   2    31    5
5   S2   3    41    7
6   S3   1    22    4
7   S3   2    32    6
8   S3   3    42    8

如您所见，数据处于商店周级别，我想计算同一周每个商店之间的欧式距离，然后取计算出的距离的平均值。因此，例如，商店S1和S2的计算如下：

    For week 1: sqrt((20-21)^2 + (2-3)^2) = sqrt(2)
    For week 2: sqrt((30-31)^2 + (4-5)^2) = sqrt(2)
    For week 3: sqrt((40-41)^2 + (6-7)^2) = sqrt(2)
    The final value for distance between S1 and S2 = sqrt(2) which is calculated as 
average distance of the 3 weeks i.e. (3 * sqrt(2)) / 3

最后输出应该如下：

   S1    S2      S3
S1 0     1.414   2.818
S2 1.414 0       some val
S3 2.818 some val 0

我对按功能分组以对数据框中的列进行分组以及对scipy.spatial.distance.cdist用于计算欧几里得距离有一些想法，但是我无法将这些概念联系起来并提出解决方案。

Answer 1

我们可以pivot然后使用numpy进行这些计算

df1  = (df.pivot(index='Store', columns='Week', values=['Sales', 'Cust_count'])
       #  .fillna(0)  # Uncomment if you want to treat missing store-weeks as 0s
       )
arr1 = df1['Sales'].to_numpy()
arr2 = df1['Cust_count'].to_numpy()

data = np.nanmean(np.sqrt(((arr1[None, :] - arr1[:, None])**2 
                         + (arr2[None, :] - arr2[:, None])**2)), 
                  axis=2)

pd.DataFrame(data, index=df1.index, columns = df1.index)

Store        S1        S2        S3
Store                              
S1     0.000000  1.414214  2.828427
S2     1.414214  0.000000  1.414214
S3     2.828427  1.414214  0.000000

Answer 2

用于permutations的循环

import itertools
s=list(itertools.permutations(df.Store.unique(), 2))
from scipy import spatial
l=[]
for x in s:
     l.append(np.sqrt(np.mean(np.sum((df[df.Store == x[0]].iloc[:, 2:].values - df[df.Store == x[1]].iloc[:, 2:].values)**2,axis=1),axis=0)))

s=pd.Series(l,index=pd.MultiIndex.from_tuples(s)).unstack()
s
Out[216]: 
          S1        S2        S3
S1       NaN  1.414214  2.828427
S2  1.414214       NaN  1.414214
S3  2.828427  1.414214       NaN

Answer 3

您可以首先在Week上merge以获得所有商店组合，然后使用欧式距离计算dist列，最后使用aggfunc='mean'计算pivot_table列：< / p>

df.merge(df, on='Week', how='left', suffixes=('','_'))\
  .assign(dist = lambda x: np.sqrt((x.Sales - x.Sales_)**2 + (x.Cust_count - x.Cust_count_)**2))\
  .pivot_table(index='Store', columns='Store_', values='dist', aggfunc='mean')

Store_        S1        S2        S3
Store                               
S1      0.000000  1.414214  2.828427
S2      1.414214  0.000000  1.414214
S3      2.828427  1.414214  0.000000

计算数据框中各组之间的欧式距离

3 个答案: