python - 每次特定列的字符串内容发生变化时从 0 重新启动索引

Question

我想在 pandas 数据框中添加一个索引列，每当特定列的字符串内容发生变化时，索引就会从 0 重新开始。

这是数据框，每次列描述中的 Take #1: 更改为 Take #2: 我想将索引重新启动为 0，当它更改为 Take #3 时也是如此:

                   Description                       A              B   
0   Take #1: (95.448; 7.399; 3.292) in A1   0.78606277704238903 0.235648
1   Take #1: (94.709; 9.619; 3.292) in A1   0.83539319038391102 0.196260
2   Take #1: (92.489; 11.839; 3.292) in A1  0.69859844446182295 0.268558
3   Take #1: (94.955; 12.825; 3.292) in A1  0.83374726772308405 0.211926
4   Take #1: (95.202; 15.538; 3.292) in A1  0.81004440784454301 0.207539
5   Take #1: (93.969; 16.031; 3.292) in A1  0.82159596681594804 0.201048
6   Take #1: (95.448; 18.744; 3.292) in A1  0.81642413139343295 0.209556
7   Take #1: (93.722; 18.991; 3.292) in A1  0.80492436885833696 0.234336
8   Take #1: (94.955; 20.964; 3.292) in A1  0.82342678308486905 0.248848
9   Take #1: (95.695; 22.937; 3.292) in A1  0.800290167331696   0.262177
10  Take #2: (95.448; 7.399; 3.292) in A1   0.77198499441146895 0.219335
11  Take #2: (94.709; 9.619; 3.292) in A1   0.83259344100952104 0.194786
12  Take #2: (92.489; 11.839; 3.292) in A1  0.53725022077560403 0.345984
13  Take #2: (94.955; 12.825; 3.292) in A1  0.84538751840591397 0.199329
14  Take #2: (95.202; 15.538; 3.292) in A1  0.83818721771240201 0.187591
15  Take #2: (93.969; 16.031; 3.292) in A1  0.81123870611190796 0.212484
16  Take #2: (95.448; 18.744; 3.292) in A1  0.79200905561447099 0.221299
17  Take #2: (93.722; 18.991; 3.292) in A1  0.81063342094421398 0.223800
18  Take #2: (94.955; 20.964; 3.292) in A1  0.81234723329544101 0.213572
19  Take #2: (95.695; 22.937; 3.292) in A1  0.82670426368713401 0.234590
20  Take #3: (95.448; 7.399; 3.292) in A1   0.80610674619674705 0.234529
21  Take #3: (94.709; 9.619; 3.292) in A1   0.84471392631530795 0.200126
22  Take #3: (92.489; 11.839; 3.292) in A1  0.55611509084701505 0.344433
23  Take #3: (94.955; 12.825; 3.292) in A1  0.82982766628265403 0.211508
24  Take #3: (95.202; 15.538; 3.292) in A1  0.82371282577514604 0.192534
25  Take #3: (93.969; 16.031; 3.292) in A1  0.81001353263855003 0.211018
26  Take #3: (95.448; 18.744; 3.292) in A1  0.80310589075088501 0.231468
27  Take #3: (93.722; 18.991; 3.292) in A1  0.813673436641693   0.234990
28  Take #3: (94.955; 20.964; 3.292) in A1  0.81725031137466397 0.212430
29  Take #3: (95.695; 22.937; 3.292) in A1  0.80750209093093905 0.230676

Answer 1

这是一种方法:

df = pd.DataFrame({'Description' : ['Take #1 : aezr', 'Take #2 : qsdkf','Take #2 : kfqdskl'],
                  'A' : [12, 46, 57], 'B' : [13, 46,57]})

# First we find where the takes # change looking in the description field
df['inter'] = [int(x[6]) for x in df['Description'] ]

# Then for each different take we fill the index column using the range function
for x in df['inter'].unique():
    mask = df['inter'] == x
    df.loc[mask, 'index'] = [x for x in range(len(df[mask]))]

# You can now drop the inter column which is no longer needed :
df.drop(columns='inter', inplace = True)

<小时/>

输出:

         Description   A   B  index
0     Take #1 : aezr  12  13    0.0
1    Take #2 : qsdkf  46  46    0.0
2  Take #2 : kfqdskl  57  57    1.0