我在熊猫中有以下数据框
1 2015_04_19_00_00_00
2 2015_04_19_01_00_00
3 2015_04_19_02_00_00
4 2015_04_19_03_00_00
5 2015_04_19_04_00_00
6 2015_04_19_05_00_00
7 2015_04_19_06_00_00
8 2020_06_10_00_00_00
9 2020_06_10_01_00_00
10 2020_06_10_02_00_00
11 2020_06_10_03_00_00
12 2020_06_10_04_00_00
13 2020_06_10_05_00_00
14 2030_04_15_01_00_00
15 2030_04_15_02_00_00
16 2030_04_15_10_00_00
17 2030_04_15_11_00_00
18 2040_05_29_01_00_00
19 2040_05_29_02_00_00
20 2040_05_29_03_00_00
21 2040_05_29_04_00_00
22 2040_05_29_05_00_00
23 2040_05_29_06_00_00
24 2040_05_29_07_00_00
25 2040_05_29_08_00_00
如何查询年份变化的指数？
最终结果应该类似于
2015 1
2020 8
2030 14
2040 18
答案 0 :(得分：4)
这是一种方式
In [148]: s = df.time_col.str.split('_').str[0]
In [149]: idx = s[s.ne(s.shift())]
In [150]: idx
Out[150]:
1 2015
8 2020
14 2030
18 2040
Name: time, dtype: object
In [151]: pd.Series(idx.index, idx.values)
Out[151]:
2015 1
2020 8
2030 14
2040 18
dtype: int64
答案 1 :(得分：1)
将duplicated与~的反布尔掩码一起使用：
a = df.col.str.split('_').str[0]
#for improve performance
#a = pd.Series([x.split('_')[0] for x in df.col], index=df.index)
b = a[~a.duplicated()]
print (b)
1 2015
8 2020
14 2030
18 2040
Name: col, dtype: object
print(pd.Series(b.index, b.values))
2015 1
2020 8
2030 14
2040 18
dtype: int64