如何获取一列特定值的第一次出现，但前提是它至少在彼此下方有 5 个连续行

Question

让我们假设我有一个熊猫数据框架，看起来像这样:

lst = [45.45454545454545, 45.45454545454545, 45.45454545454545, 45.45454545454545, 45.45454545454545, 36.36363636363637, 36.36363636363637, 36.36363636363637, 27.27272727272727, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 27.27272727272727, 0.0, 0.0, 27.27272727272727, 0.0, 0.0, 0.0, 0.0, 27.27272727272727, 0.0, 0.0, 0.0, 36.36363636363637, 0.0, 27.27272727272727, 0.0, 27.27272727272727, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 27.27272727272727, 27.27272727272727, 54.54545454545454, 27.27272727272727, 36.36363636363637, 36.36363636363637, 54.54545454545454, 36.36363636363637, 45.45454545454545, 45.45454545454545, 36.36363636363637, 36.36363636363637, 45.45454545454545, 45.45454545454545, 36.36363636363637, 45.45454545454545, 36.36363636363637, 45.45454545454545, 36.36363636363637, 45.45454545454545, 36.36363636363637, 36.36363636363637, 36.36363636363637, 0.0, 36.36363636363637, 27.27272727272727, 0.0, 36.36363636363637, 0.0, 36.36363636363637, 36.36363636363637, 0.0, 0.0, 27.27272727272727, 0.0, 36.36363636363637, 0.0, 0.0, 0.0, 0.0, 36.36363636363637, 36.36363636363637, 0.0, 36.36363636363637, 36.36363636363637, 27.27272727272727, 27.27272727272727, 36.36363636363637, 36.36363636363637, 36.36363636363637, 36.36363636363637, 0.0, 27.27272727272727, 0.0, 0.0, 0.0, 27.27272727272727, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 27.27272727272727, 36.36363636363637, 0.0, 0.0, 0.0, 0.0, 0.0]
df = pd.DataFrame(lst,columns =['%'])
df.index.name='Time/ps'
df

现在，我想知道第一次出现在什么时间?"%";减小到0.0但仅条件是0(0)存在至少在彼此下面连续5行。我试着用这段代码来做它，它部分工作:

for k, v in df[df['%'] == 0.000000].groupby((df['%'] != 0.000000).cumsum()):
print(f'[group {k}]')
print(v)
print('n')

然而，问题是，我不知道如何获得信息时，有0.0's在"%"列下至少5个连续的行。这段代码打印所有出现的情况，我可以滚动它，但我想自动完成。我想要的输出看起来像这样:Time/ps: 9

谢谢你的建议

Answer 1

天真的答案可能是将您的代码像这样做:

for k, v in df[df['%'] == 0.000000].groupby((df['%'] != 0.000000).cumsum()):
if len(v) > 5:
print("Time/ps:", k)
break

一个更好的方式可能是这样的:

df[df['%'] == 0.000000].groupby((df['%'] != 0.000000).cumsum()).filter(lambda x: len(x) > 5)

我给你分组代码，然后使用过滤器过滤掉长度小于5的组。

这给出了这个数据帧:

Time/ps     
9       0.0
10      0.0
11      0.0
12      0.0
13      0.0
14      0.0
15      0.0
16      0.0
17      0.0
35      0.0
36      0.0
37      0.0
38      0.0
39      0.0
40      0.0
41      0.0
99      0.0
100     0.0
101     0.0
102     0.0
103     0.0
104     0.0
105     0.0
106     0.0

你说你想要第一次出现:

index = df[df['%'] == 0.000000].groupby((df['%'] != 0.000000).cumsum()).filter(lambda x: len(x) > 5).iloc[0].name
print('Time/ps:', index)
# Time/ps: 9

Answer 2

如果所有值都是非负的，则可以使用长度为5的滚动窗口和，并使用argmin找到第一个零:

k = 5
df.index[df['%'].rolling(k).sum().argmin() - k + 1]

(如有负值，需先做.abs()，后做rolling())

Answer 3

import pandas as pd
import numpy as np
number = 0.0 #45.45454545454545
df = pd.DataFrame(lst, columns=['p']) #use more convenient name of column
df['dif'] = np.abs(df.p.diff(1)) + np.abs(df.p.diff(2)) + np.abs(df.p.diff(3)) + np.abs(df.p.diff(4))
positions = df.index[(df.p==number) & (df.dif==0.0)]
first_index = positions[0] - 4