在熊猫中只删除连续重复的最有效方法是什么？
drop_duplicates给出了这个:
In [3]: a = pandas.Series([1,2,2,3,2], index=[1,2,3,4,5])
In [4]: a.drop_duplicates()
Out[4]:
1 1
2 2
4 3
dtype: int64
但我想要这个:
In [4]: a.something()
Out[4]:
1 1
2 2
4 3
5 2
dtype: int64
EdC*_*ica 72
用途shift:
a.loc[a.shift(-1) != a]
Out[3]:
1 1
3 2
4 3
5 2
dtype: int64
所以上面使用布尔标准,我们将数据帧与移位-1行的数据帧进行比较以创建掩码
另一种方法是使用diff:
In [82]:
a.loc[a.diff() != 0]
Out[82]:
1 1
2 2
4 3
5 2
dtype: int64
但是如果您有大量行,这比原始方法慢.
更新
感谢Bjarke Ebert指出一个微妙的错误,我实际应该使用shift(1)或者只是shift()因为默认值是1的周期,这将返回第一个连续值:
In [87]:
a.loc[a.shift() != a]
Out[87]:
1 1
2 2
4 3
5 2
dtype: int64
注意索引值的差异,谢谢@BjarkeEbert!
joh*_*135 10
这是一个更新,它将使其适用于多个列.使用".any(axis = 1)"组合每列的结果:
cols = ["col1","col2","col3"]
de_dup = a[cols].loc[(a[cols].shift() != a[cols]).any(axis=1)]
由于我们追求的是most efficient way性能，因此让我们使用数组数据来利用 NumPy。我们将切片一次性切片并进行比较，类似于前面讨论的移位方法@EdChum's post。但是使用 NumPy 切片我们最终会得到一个无一个数组，所以我们需要True在开始时连接一个元素来选择第一个元素，因此我们会有一个像这样的实现 -
def drop_consecutive_duplicates(a):
ar = a.values
return a[np.concatenate(([True],ar[:-1]!= ar[1:]))]
样品运行 -
In [149]: a
Out[149]:
1 1
2 2
3 2
4 3
5 2
dtype: int64
In [150]: drop_consecutive_duplicates(a)
Out[150]:
1 1
2 2
4 3
5 2
dtype: int64
大型阵列的时序比较@EdChum's solution-
In [142]: a = pd.Series(np.random.randint(1,5,(1000000)))
In [143]: %timeit a.loc[a.shift() != a]
100 loops, best of 3: 12.1 ms per loop
In [144]: %timeit drop_consecutive_duplicates(a)
100 loops, best of 3: 11 ms per loop
In [145]: a = pd.Series(np.random.randint(1,5,(10000000)))
In [146]: %timeit a.loc[a.shift() != a]
10 loops, best of 3: 136 ms per loop
In [147]: %timeit drop_consecutive_duplicates(a)
10 loops, best of 3: 114 ms per loop
所以，有一些改进！
只为价值获得重大提升！
如果只需要值，我们可以通过简单地索引到数组数据来获得重大提升，就像这样 -
def drop_consecutive_duplicates(a):
ar = a.values
return ar[np.concatenate(([True],ar[:-1]!= ar[1:]))]
样品运行 -
In [170]: a = pandas.Series([1,2,2,3,2], index=[1,2,3,4,5])
In [171]: drop_consecutive_duplicates(a)
Out[171]: array([1, 2, 3, 2])
时间——
In [173]: a = pd.Series(np.random.randint(1,5,(10000000)))
In [174]: %timeit a.loc[a.shift() != a]
10 loops, best of 3: 137 ms per loop
In [175]: %timeit drop_consecutive_duplicates(a)
10 loops, best of 3: 61.3 ms per loop
pd.Series这是一个处理和 的函数pd.Dataframes。您可以屏蔽/删除，选择轴，最后选择使用“任意”或“全部”“NaN”删除。它没有在计算时间方面进行优化，但它的优点是健壮且非常清晰。
import numpy as np
import pandas as pd
# To mask/drop successive values in pandas
def Mask_Or_Drop_Successive_Identical_Values(df, drop=False,
keep_first=True,
axis=0, how='all'):
'''
#Function built with the help of:
# 1) /sf/ask/3389972141/
# 2) /sf/ask/1362478981/
Input:
df should be a pandas.DataFrame of a a pandas.Series
Output:
df of ts with masked or dropped values
'''
# Mask keeping the first occurrence
if keep_first:
df = df.mask(df.shift(1) == df)
# Mask including the first occurrence
else:
df = df.mask((df.shift(1) == df) | (df.shift(-1) == df))
# Drop the values (e.g. rows are deleted)
if drop:
return df.dropna(axis=axis, how=how)
# Only mask the values (e.g. become 'NaN')
else:
return df
以下是要包含在脚本中的测试代码：
if __name__ == "__main__":
# With time series
print("With time series:\n")
ts = pd.Series([1,1,2,2,3,2,6,6,float('nan'), 6,6,float('nan'),float('nan')],
index=[0,1,2,3,4,5,6,7,8,9,10,11,12])
print("#Original ts:")
print(ts)
print("\n## 1) Mask keeping the first occurrence:")
print(Mask_Or_Drop_Successive_Identical_Values(ts, drop=False,
keep_first=True))
print("\n## 2) Mask including the first occurrence:")
print(Mask_Or_Drop_Successive_Identical_Values(ts, drop=False,
keep_first=False))
print("\n## 3) Drop keeping the first occurrence:")
print(Mask_Or_Drop_Successive_Identical_Values(ts, drop=True,
keep_first=True))
print("\n## 4) Drop including the first occurrence:")
print(Mask_Or_Drop_Successive_Identical_Values(ts, drop=True,
keep_first=False))
# With dataframes
print("With dataframe:\n")
df = pd.DataFrame(np.random.randn(15, 3))
df.iloc[4:9,0]=40
df.iloc[8:15,1]=22
df.iloc[8:12,2]=0.23
print("#Original df:")
print(df)
print("\n## 5) Mask keeping the first occurrence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=False,
keep_first=True))
print("\n## 6) Mask including the first occurrence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=False,
keep_first=False))
print("\n## 7) Drop 'any' keeping the first occurrence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=True,
keep_first=True,
how='any'))
print("\n## 8) Drop 'all' keeping the first occurrence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=True,
keep_first=True,
how='all'))
print("\n## 9) Drop 'any' including the first occurrence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=True,
keep_first=False,
how='any'))
print("\n## 10) Drop 'all' including the first occurrence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=True,
keep_first=False,
how='all'))
这是预期的结果：
With time series:
#Original ts:
0 1.0
1 1.0
2 2.0
3 2.0
4 3.0
5 2.0
6 6.0
7 6.0
8 NaN
9 6.0
10 6.0
11 NaN
12 NaN
dtype: float64
## 1) Mask keeping the first occurrence:
0 1.0
1 NaN
2 2.0
3 NaN
4 3.0
5 2.0
6 6.0
7 NaN
8 NaN
9 6.0
10 NaN
11 NaN
12 NaN
dtype: float64
## 2) Mask including the first occurrence:
0 NaN
1 NaN
2 NaN
3 NaN
4 3.0
5 2.0
6 NaN
7 NaN
8 NaN
9 NaN
10 NaN
11 NaN
12 NaN
dtype: float64
## 3) Drop keeping the first occurrence:
0 1.0
2 2.0
4 3.0
5 2.0
6 6.0
9 6.0
dtype: float64
## 4) Drop including the first occurrence:
4 3.0
5 2.0
dtype: float64
With dataframe:
#Original df:
0 1 2
0 -1.890137 -3.125224 -1.029065
1 -0.224712 -0.194742 1.891365
2 1.009388 0.589445 0.927405
3 0.212746 -0.392314 -0.781851
4 40.000000 1.889781 -1.394573
5 40.000000 -0.470958 -0.339213
6 40.000000 1.613524 0.271641
7 40.000000 -1.810958 -1.568372
8 40.000000 22.000000 0.230000
9 -0.296557 22.000000 0.230000
10 -0.921238 22.000000 0.230000
11 -0.170195 22.000000 0.230000
12 1.460457 22.000000 -0.295418
13 0.307825 22.000000 -0.759131
14 0.287392 22.000000 0.378315
## 5) Mask keeping the first occurrence:
0 1 2
0 -1.890137 -3.125224 -1.029065
1 -0.224712 -0.194742 1.891365
2 1.009388 0.589445 0.927405
3 0.212746 -0.392314 -0.781851
4 40.000000 1.889781 -1.394573
5 NaN -0.470958 -0.339213
6 NaN 1.613524 0.271641
7 NaN -1.810958 -1.568372
8 NaN 22.000000 0.230000
9 -0.296557 NaN NaN
10 -0.921238 NaN NaN
11 -0.170195 NaN NaN
12 1.460457 NaN -0.295418
13 0.307825 NaN -0.759131
14 0.287392 NaN 0.378315
## 6) Mask including the first occurrence:
0 1 2
0 -1.890137 -3.125224 -1.029065
1 -0.224712 -0.194742 1.891365
2 1.009388 0.589445 0.927405
3 0.212746 -0.392314 -0.781851
4 NaN 1.889781 -1.394573
5 NaN -0.470958 -0.339213
6 NaN 1.613524 0.271641
7 NaN -1.810958 -1.568372
8 NaN NaN NaN
9 -0.296557 NaN NaN
10 -0.921238 NaN NaN
11 -0.170195 NaN NaN
12 1.460457 NaN -0.295418
13 0.307825 NaN -0.759131
14 0.287392 NaN 0.378315
## 7) Drop 'any' keeping the first occurrence:
0 1 2
0 -1.890137 -3.125224 -1.029065
1 -0.224712 -0.194742 1.891365
2 1.009388 0.589445 0.927405
3 0.212746 -0.392314 -0.781851
4 40.000000 1.889781 -1.394573
## 8) Drop 'all' keeping the first occurrence:
0 1 2
0 -1.890137 -3.125224 -1.029065
1 -0.224712 -0.194742 1.891365
2 1.009388 0.589445 0.927405
3 0.212746 -0.392314 -0.781851
4 40.000000 1.889781 -1.394573
5 NaN -0.470958 -0.339213
6 NaN 1.613524 0.271641
7 NaN -1.810958 -1.568372
8 NaN 22.000000 0.230000
9 -0.296557 NaN NaN
10 -0.921238 NaN NaN
11 -0.170195 NaN NaN
12 1.460457 NaN -0.295418
13 0.307825 NaN -0.759131
14 0.287392 NaN 0.378315
## 9) Drop 'any' including the first occurrence:
0 1 2
0 -1.890137 -3.125224 -1.029065
1 -0.224712 -0.194742 1.891365
2 1.009388 0.589445 0.927405
3 0.212746 -0.392314 -0.781851
## 10) Drop 'all' including the first occurrence:
0 1 2
0 -1.890137 -3.125224 -1.029065
1 -0.224712 -0.194742 1.891365
2 1.009388 0.589445 0.927405
3 0.212746 -0.392314 -0.781851
4 NaN 1.889781 -1.394573
5 NaN -0.470958 -0.339213
6 NaN 1.613524 0.271641
7 NaN -1.810958 -1.568372
9 -0.296557 NaN NaN
10 -0.921238 NaN NaN
11 -0.170195 NaN NaN
12 1.460457 NaN -0.295418
13 0.307825 NaN -0.759131
14 0.287392 NaN 0.378315
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