Question

对于以下df，我想计算列Inst_Dist的累积总和并另存为Cumu_Dist，而WDir_Deg的值保持不变。当WDir_Deg中的值更改时，我需要重新启动累积和。

因此

index | WDir_Deg | Inst_Dist | Cumu_Dist
0     | 289      | 20        | NaN
1     | 285      | 17        | NaN
2     | 285      | 19        | NaN
3     | 287      | 19        | NaN
4     | 289      | 10        | NaN

成为

index | WDir_Deg | Inst_Dist | Cumu_Dist
0     | 289      | 20        | 20
1     | 285      | 17        | 17
2     | 285      | 19        | 36
3     | 287      | 19        | 19
4     | 289      | 10        | 10

我的非惯用（极慢）Python代码如下。如果有人可以指导我如何使代码更快且更惯用，我将不胜感激。

prev_angle = -1
curr_cumu_dist = 0
for curr_ind in df.index:
    curr_angle = df.loc[curr_ind, 'WDir_Deg']
    if prev_angle == curr_angle:
        curr_cumu_dist += df.loc[curr_ind, 'Inst_Dist']
        df.loc[curr_ind, 'Cumu_Dist'] = curr_cumu_dist
    else:
        prev_angle = curr_angle
        curr_cumu_dist = df.loc[curr_ind, 'Inst_Dist']
        df.loc[curr_ind, 'Cumu_Dist'] = curr_cumu_dist

Answer 1

有点棘手。引用此问题/答案Pandas groupby cumulative sum

我提出了解决方案

df['Cumu_Dist'] = df.groupby('WDir_Deg').Inst_Dist.cumsum()

返回哪个

   index  WDir_Deg  Inst_Dist  Cumu_Dist
0      0       285         17         17
1      1       285         19         36
2      2       287         19         19
3      3       289         20         20

这使用pandas版0.23.4

Answer 2

对连续的组使用不等于ne，shift和cumsum的带有比较Series列的助手WDir_Deg并将其传递给{{3} }：

s = df['WDir_Deg'].ne(df['WDir_Deg'].shift()).cumsum()
df['Cumu_Dist'] = df.groupby(s)['Inst_Dist'].cumsum()
print (df)
   WDir_Deg  Inst_Dist  Cumu_Dist
0       289         20         20
1       285         17         17
2       285         19         36
3       287         19         19
4       289         10         10

详细信息：

print (s)
0    1
1    2
2    2
3    3
4    4
Name: WDir_Deg, dtype: int32

在另一列的值保持不变的情况下计算累计总和

2 个答案: