矩阵的标题

Question

在以下数据集中，我需要添加列和行，以便我知道例如员工“12”从雇主“a”到雇主“b”的位置。这是我的数据集employerEmployeeEdges = [(12, 'a'), (15, 'a'), (17, 'a'), (12, 'a'), (15, 'a'), (23, 'b'), (12, 'b'), (18, 'b'), (12, 'b'), (12, 'b'), (15, 'a'), (12, 'a'), (15, 'a'), (15, 'a'), (24, 'c'), (12, 'c')]employerEmployeeEdges=np.array(employerEmployeeEdges)#print(employerEmployeeEdges)unique_employee = np.unique(employerEmployeeEdges[:,1])n_unique = len(unique_employee)#print(unique_employee)Q = np.zeros([n_unique,n_unique])for n, employer_employee in enumerate(employerEmployeeEdges):    #print(employer_employee)    #copy the array for the original o be intact    eee = np.copy(employerEmployeeEdges)    #sustitue the current tuple with a empty one to avoid self comparing    eee[n] = (None,None)    #get the index for the current employee, the one on the y axis    employee_index = np.where(employer_employee[1] == unique_employee)    #get the indexes where the the employees letter match    eq_index = np.where(eee[:,0] == employer_employee[0])[0]    eq_employee = eee[eq_index,1]    #add at the final array Q by index    for emp in eq_employee:        print(np.unique(emp))        emp_index = np.where(unique_employee == emp)        #print(emp)        Q[employee_index,emp_index]+= 1        #df = pd.DataFrame(Q, columns=emp, index=emp)print(Q) [[26.  9.  3.] [ 9.  6.  3.] [ 3.  3.  0.]]我想在上面的这个矩阵中添加列和行标题这是我到目前为止所做的：for index, row in enumerate(Q):    if index < len(Q)-1:        print('{}\t'.format(str(index + 1))),    else:        print(' \t'),    print('|'.join('{0:.2f}'.format(x) for x in row))1   26.00|9.00|3.002   9.00|6.00|3.00    3.00|3.00|0.00由于某种原因，我无法向数组添加列或行。我需要做什么？这个数组应该看起来像（我想要的输出）       a    b    ca   26.00|9.00|3.00b   9.00|6.00|3.00b   3.00|3.00|0.00基于安德鲁的帮助，这里是解决方案df = pd.DataFrame(Q)df.index = unique_employeedf.columns = unique_employeeprint(df)      a    b    ca  26.0  9.0  3.0b   9.0  6.0  3.0c   3.0  3.0  0.0

Answer 1

您可以使用熊猫并指定index（行标签）和columns（列标签）来匹配您的unique_employee数组。

import pandas as pd 

print(Q) 

[[26.  9.  3.]

 [ 9.  6.  3.]

 [ 3.  3.  0.]]

df = pd.DataFrame(Q)

df.index = unique_employee

df.columns = unique_employee

print(df)

      a    b    c

a  26.0  9.0  3.0

b   9.0  6.0  3.0

c   3.0  3.0  0.0