计算熊猫循环中的缺失值

Question

• 如何在Pandas中运行循环，该循环返回包含零值的缺失值的列表？

循环应使用full_scores_list中的值，检查该值是否在Home_team_scores列中。 如果这样，则计算并输出频率，否则返回零。

我的名单很长，在两个赛季中一支球队的得分不同。

       My code is:
       all_scores = []
       for score in range(0,len(full_scores_list)):
           if full_scores_list[score] == '0 0':
                    all_scores.append(data1.Home_team_scores.value_counts()['0 0'])
           elif full_scores_list[score] == '0 1':
                   all_scores.append(data1.Home_team_scores.value_counts()['0     1'])
           elif full_scores_list[score] == '0 2':
                all_scores.append(data1.Home_team_scores.value_counts()['0 2']) 
               all_scores.append(data1.Home_team_scores.value_counts()['0 2'])
           elif full_scores_list[score] == '0 3':
               all_scores.append(data1.Home_team_scores.value_counts()['0 3'])
           elif full_scores_list[score] == '0 4':
               all_scores.append(data1.Home_team_scores.value_counts()['0 4'])
           elif full_scores_list[score] == '0 5':
               all_scores.append(data1.Home_team_scores.value_counts()['0 5'])  
           if full_scores_list[score] == '1 0':
                all_scores.append(data1.Home_team_scores.value_counts()['1 0'])
           elif full_scores_list[score] == '1 1':
               all_scores.append(data1.Home_team_scores.value_counts()['1 1'])
           elif full_scores_list[score] == '1 2':
               all_scores.append(data1.Home_team_scores.value_counts()['1 2'])
           elif full_scores_list[score] == '1 3':
               all_scores.append(data1.Home_team_scores.value_counts()['1 3'])
           elif full_scores_list[score] == '1 4':
               all_scores.append(data1.Home_team_scores.value_counts()['1 4'])
           elif full_scores_list[score] == '1 5':
               all_scores.append(data1.Home_team_scores.value_counts()['1 5'])


       all_scores

Answer 1

大致情况如何：

com_vals = df['Home_team_scores'].unique()
df['full_scores_list'].apply(lambda v: v in com_vals)

根据评论改进：

除了lambda函数，您可以对apply()使用辅助函数：

com_vals = df['Home_team_scores'].unique()
def helper():
  return v in com_vals
df['full_scores_list'].apply(helper)

您可以使用helper()函数对输出和条件进行更细粒度的控制。