pandas 在pd.dataframe中寻找一组连续常数值的简洁的Python方法是什么？

carvr3hs 于 2022-12-16 发布在 Python

关注(0)|答案(2)|浏览(171)

我正在寻找一种方法来标记连续的常量值（例如n），这些值在pd. datafame（例如df）中连续为常量。
我已经写了一些代码，如果一个值与n/2个下一个和n/2个前一个数据点的差值为零，那么这个值将被标记。

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。
我期待的是：

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

pandas

来源：https://stackoverflow.com/questions/55759947/what-is-a-clean-pythonic-method-to-find-a-set-of-sequential-constant-values-in-a

2条答案

按热度按时间

xmq68pz91#

IIUC，使用DataFrame.groupby，石斑鱼为Series.diff、.ne(0)，然后为.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

我们分组所依据的系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

当你使用transform返回一个与原始DataFrame形状相同的对象，并聚合为“size”时，你会得到：

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，将Series.ge（>=）与您的值n进行简单比较

赞(0）回复(0）举报 2022-12-16

voase2hg2#

参见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

赞(0）回复(0）举报 2022-12-16

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pandas 在pd.dataframe中寻找一组连续常数值的简洁的Python方法是什么？

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