什么是在 pd.dataframe 中查找一组顺序常量值的干净 pythonic 方法

Question

我正在寻找一种方法来标记pd.dataframe(例如df(中连续常量值(例如n(，这些值是常量。

我编写了一些代码，如果值与下一个 n/2 和前一个数据点的 n/2 之差为零，则通过这些代码将标记该值。

n = 5         # the minimum number of sequential constant values  
#to create a adatframe example
df=pd.DataFrame(np.random.randn(25), index=pd.date_range(start='2010-1-1',end='2010-1-2',freq='H'), columns=['value'])
#to modify the dataframe to have several sets of constant values
df[1:10]=23
df[20:26]=10
df[15:17]=15
for i in np.arange(1, int(n/2)):
    # to calcualte the difference between value and ith previous values
    df_diff['delta_' + str(i)] = (df['value'].diff(periods=i)).abs()
    # to calcualte the difference between value and ith next values
    df_diff['delta_' + str(-i)] = (df['value'].diff(periods=-i)).abs()
# to filter the results (e.g. as a boolean)
result_1 = (df_diff[:] <= 0).all(axis=1)
result_2 = (df_diff[:] <= 0).any(axis=1)

此示例中的result_1和results_2未提供正确答案。

我的期望是：

2010-01-01 00:00:00    False
2010-01-01 01:00:00     True
2010-01-01 02:00:00     True
2010-01-01 03:00:00     True
2010-01-01 04:00:00     True
2010-01-01 05:00:00     True
2010-01-01 06:00:00     True
2010-01-01 07:00:00     True
2010-01-01 08:00:00     True
2010-01-01 09:00:00     True
2010-01-01 10:00:00    False
2010-01-01 11:00:00    False
2010-01-01 12:00:00    False
2010-01-01 13:00:00    False
2010-01-01 14:00:00    False
2010-01-01 15:00:00    False
2010-01-01 16:00:00    False
2010-01-01 17:00:00    False
2010-01-01 18:00:00    False
2010-01-01 19:00:00    False
2010-01-01 20:00:00     True
2010-01-01 21:00:00     True
2010-01-01 22:00:00     True
2010-01-01 23:00:00     True
2010-01-02 00:00:00     True

Answer 1

IIUC，使用DataFrame.groupby石斑鱼Series.diff，.ne(0)然后.cumsum：

df.groupby(df.value.diff().ne(0).cumsum())['value'].transform('size').ge(n)

[出]

2010-01-01 00:00:00    False
2010-01-01 01:00:00     True
2010-01-01 02:00:00     True
2010-01-01 03:00:00     True
2010-01-01 04:00:00     True
2010-01-01 05:00:00     True
2010-01-01 06:00:00     True
2010-01-01 07:00:00     True
2010-01-01 08:00:00     True
2010-01-01 09:00:00     True
2010-01-01 10:00:00    False
2010-01-01 11:00:00    False
2010-01-01 12:00:00    False
2010-01-01 13:00:00    False
2010-01-01 14:00:00    False
2010-01-01 15:00:00    False
2010-01-01 16:00:00    False
2010-01-01 17:00:00    False
2010-01-01 18:00:00    False
2010-01-01 19:00:00    False
2010-01-01 20:00:00     True
2010-01-01 21:00:00     True
2010-01-01 22:00:00     True
2010-01-01 23:00:00     True
2010-01-02 00:00:00     True
Freq: H, Name: value, dtype: bool

解释

我们分组的序列将是具有相等值的连续组：

s = df.value.diff().ne(0).cumsum()
2010-01-01 00:00:00     1
2010-01-01 01:00:00     2
2010-01-01 02:00:00     2
2010-01-01 03:00:00     2
2010-01-01 04:00:00     2
2010-01-01 05:00:00     2
2010-01-01 06:00:00     2
2010-01-01 07:00:00     2
2010-01-01 08:00:00     2
2010-01-01 09:00:00     2
2010-01-01 10:00:00     3
2010-01-01 11:00:00     4
2010-01-01 12:00:00     5
2010-01-01 13:00:00     6
2010-01-01 14:00:00     7
2010-01-01 15:00:00     8
2010-01-01 16:00:00     8
2010-01-01 17:00:00     9
2010-01-01 18:00:00    10
2010-01-01 19:00:00    11
2010-01-01 20:00:00    12
2010-01-01 21:00:00    12
2010-01-01 22:00:00    12
2010-01-01 23:00:00    12
2010-01-02 00:00:00    12
Freq: H, Name: value, dtype: int32

当您按这些"组 id"分组时，使用 transform 返回与原始DataFrame相同的形状的对象，聚合到"大小"，您将获得：

s.groupby(s).transform('size')
2010-01-01 00:00:00    1
2010-01-01 01:00:00    9
2010-01-01 02:00:00    9
2010-01-01 03:00:00    9
2010-01-01 04:00:00    9
2010-01-01 05:00:00    9
2010-01-01 06:00:00    9
2010-01-01 07:00:00    9
2010-01-01 08:00:00    9
2010-01-01 09:00:00    9
2010-01-01 10:00:00    1
2010-01-01 11:00:00    1
2010-01-01 12:00:00    1
2010-01-01 13:00:00    1
2010-01-01 14:00:00    1
2010-01-01 15:00:00    2
2010-01-01 16:00:00    2
2010-01-01 17:00:00    1
2010-01-01 18:00:00    1
2010-01-01 19:00:00    1
2010-01-01 20:00:00    5
2010-01-01 21:00:00    5
2010-01-01 22:00:00    5
2010-01-01 23:00:00    5
2010-01-02 00:00:00    5
Freq: H, Name: value, dtype: int64

从这里开始，这是一个简单的Series.ge(>=(与您的价值n比较

Answer 2

请参阅 https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.duplicated.html

import pandas as pd

df = pd.DataFrame({
    'brand': ['Yum Yum', 'Yum Yum', 'Indomie', 'Indomie', 'Indomie'],
    'style': ['cup', 'cup', 'cup', 'pack', 'pack'],
    'rating': [4, 4, 3.5, 15, 5]
})
df
#    brand     style   rating
# 0  Yum Yum   cup     4.0
# 1  Yum Yum   cup     4.0
# 2  Indomie   cup     3.5
# 3  Indomie   pack    15.0
# 4  Indomie   pack     5.0
df.duplicated(keep=False)
# 0     True
# 1     True
# 2     False
# 3     False
# 4     False
# dtype: bool