我有以下数据帧,其中对value
列进行了排序:
df = pd.DataFrame({'variable': {0: 'Chi', 1: 'San Antonio', 2: 'Dallas', 3: 'PHL', 4: 'Houston', 5: 'NY', 6: 'Phoenix', 7: 'San Diego', 8: 'LA', 9: 'San Jose', 10: 'SF'}, 'value': {0: 191.28, 1: 262.53, 2: 280.21, 3: 283.08, 4: 290.75, 5: 295.72, 6: 305.6, 7: 357.89, 8: 380.07, 9: 452.71, 10: 477.67}})
输出:
variable value
0 Chi 191.28
1 San Antonio 262.53
2 Dallas 280.21
3 PHL 283.08
4 Houston 290.75
5 NY 295.72
6 Phoenix 305.60
7 San Diego 357.89
8 LA 380.07
9 San Jose 452.71
10 SF 477.67
我想找到相邻值之间的距离小于10的值:
df['value'].diff() < 10
输出:
0 False
1 False
2 False
3 True
4 True
5 True
6 True
7 False
8 False
9 False
10 False
Name: value, dtype: bool
现在,我想将彼此太近的True
值平均隔开。这个想法是在True
序列(280.21)之前取第一个值,然后向每个下一个True
值(累加和)加5:第一个True
= 280 + 5,第二个True
= 280 + 5 + 5,第三个True
= 280 + 5 + 5 ...
预期输出:
variable value
0 Chi 191.28
1 San Antonio 262.53
2 Dallas 280.21
3 PHL 285.21 <-
4 Houston 290.21 <-
5 NY 295.21 <-
6 Phoenix 300.21 <-
7 San Diego 357.89
8 LA 380.07
9 San Jose 452.71
10 SF 477.67
我的解决方案:
mask = df['value'].diff() < 10
df.loc[mask, 'value'] = 5
df.loc[mask | mask.shift(-1), 'value'] = last_day[mask | mask.shift(-1), 'value'].cumsum()
也许是更优雅的一种。
让我们尝试一下:
df = pd.DataFrame({'variable': {0: 'Chi', 1: 'San Antonio', 2: 'Dallas', 3: 'PHL', 4: 'Houston', 5: 'NY', 6: 'Phoenix', 7: 'San Diego', 8: 'LA', 9: 'San Jose', 10: 'SF'}, 'value': {0: 191.28, 1: 262.53, 2: 280.21, 3: 283.08, 4: 290.75, 5: 295.72, 6: 305.6, 7: 357.89, 8: 380.07, 9: 452.71, 10: 477.67}})
s = df['value'].diff() < 10
add_amt = s.cumsum().mask(~s) * 5
df_out = df.assign(value=df['value'].mask(add_amt.notna()).ffill() + add_amt.fillna(0))
df_out
输出:
variable value
0 Chi 191.28
1 San Antonio 262.53
2 Dallas 280.21
3 PHL 285.21
4 Houston 290.21
5 NY 295.21
6 Phoenix 300.21
7 San Diego 357.89
8 LA 380.07
9 San Jose 452.71
10 SF 477.67