import pandas as pd
num_pos = [(num,pos) for i in data for pos,num in enumerate(i)]
df = pd.DataFrame(num_pos,columns = ['number','position']).assign(numoftimesinposition = 1)
df = df.astype(int).groupby(['number','position']).count().reset_index()

df1 = df.groupby('number').numoftimesinposition.sum().reset_index().\
    rename(columns = {'numoftimesinposition':'numofoccurences'}).\
    merge(df, on='number')

print(df1)
    number  numofoccurences  position  numoftimesinposition
0        1                2         4                     1
1        1                2         5                     1
4        2                1         0                     1
7        4                1         3                     1
9        6                1         2                     1
2       12                1         4                     1
3       13                1         0                     1
5       22                1         0                     1
6       23                1         3                     1
8       45                2         1                     2
10      65                3         1                     1
11      65                3         5                     2
12      67                1         2                     1
13      72                1         3                     1
14      98                2         2                     1
15      98                2         4                     1

如果感觉上面的代码很慢，那么使用collections中的Counter：

import pandas as pd
from collections import Counter

num_pos = [(int(num),pos) for i in data for pos,num in enumerate(i)]

count_data = [(num,pos,occurence) for (num,pos), occurence in Counter(num_pos).items()]

df = pd.DataFrame(count_data, columns = ['num','pos','occurence']).sort_values(by='num')

df['total_occurence'] = [Counter(df.num).get(num) for num in df.num]
print(df)

这应该可以解决您的查询（应该比非常慢的groupby（您需要两个）和其他Pandas操作更快）-

#get the list of lists into a 2d numpy array
dd = np.array(data).astype(int)

#get vocab of all unique numbers
vocab = np.unique(dd.flatten())

#loop thru vocab and get sum of occurances in each index position
df = pd.DataFrame([[i]+list(np.sum((dd==i).astype(int), axis=0)) for i in vocab])

#rename cols
df.columns = ['num', 0, 1, 2, 3, 4, 5] 

#create total occurances of the item
df['numoccurances'] = df.iloc[:,1:].sum(axis=1)  
 
#Stack the position counts and rename cols
stats = pd.DataFrame(df.set_index(['num','numoccurances']).\
                     stack()).reset_index().\
                     set_axis(['num', 'numoccurances', 'position', 'numtimesinposition'], axis=1)

#get only rows with occurances
stats = stats[stats['numtimesinposition']>0].reset_index(drop=True) 
stats

num  numoccurances  position  numtimesinposition
0     1              2         4                   1
1     1              2         5                   1
2     2              1         0                   1
3     4              1         3                   1
4     6              1         2                   1
5    12              1         4                   1
6    13              1         0                   1
7    22              1         0                   1
8    23              1         3                   1
9    45              2         1                   2
10   65              3         1                   1
11   65              3         5                   2
12   67              1         2                   1
13   72              1         3                   1
14   98              2         2                   1
15   98              2         4                   1

结果显示-
1在您共享的样本数据中总共出现了2次，在第5和第6位各出现了1次。同样，2总共出现了1次，在第1位也是如此。