Python 熊猫：删除连续的重复项
在大熊猫身上，最有效的方法是什么 drop_duplicates提供了以下功能：Python 熊猫：删除连续的重复项,python,pandas,Python,Pandas,在大熊猫身上，最有效的方法是什么 drop_duplicates提供了以下功能： 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: int6
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
In [4]: a.something()
Out[4]:
1 1
2 2
4 3
5 2
dtype: int64
shift（1）shift（）In [87]:
a.loc[a.shift() != a]
Out[87]:
1 1
2 2
4 3
5 2
dtype: int64
请注意索引值的差异，谢谢@BjarkeEbert 这里有一个更新，它可以处理多个列。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
最有效的方法@EdChum的帖子中讨论的移位方法。但是使用NumPy切片，我们最终会少得到一个数组，因此我们需要在开始时连接一个True
元素来选择第一个元素，因此我们将有一个这样的实现-
def drop_consecutive_duplicates(a):
ar = a.values
return a[np.concatenate(([True],ar[:-1]!= ar[1:]))]
def drop_consecutive_duplicates(a):
ar = a.values
return ar[np.concatenate(([True],ar[:-1]!= ar[1:]))]
样本运行-
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
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])
大型阵列上的计时比较-
所以，有一些改进
仅为价值观获得重大提升
如果只需要这些值，我们可以通过简单地索引到数组数据中来获得巨大的提升，就像这样-
def drop_consecutive_duplicates(a):
ar = a.values
return a[np.concatenate(([True],ar[:-1]!= ar[1:]))]
def drop_consecutive_duplicates(a):
ar = a.values
return ar[np.concatenate(([True],ar[:-1]!= ar[1:]))]
样本运行-
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
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])
时间安排-
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
对于其他堆栈探索者，根据上面的johnml1135答案进行构建。这将从多个列中删除下一个副本，但不会删除所有列。当对数据帧进行排序时，它将保留第一行，但如果“cols”匹配，则删除第二行，即使有更多的列具有不匹配的信息
cols = ["col1","col2","col3"]
df = df.loc[(df[cols].shift() != df[cols]).any(axis=1)]
这是一个处理pd.Series
和pd.Dataframes
的函数。您可以屏蔽/删除，选择轴，最后选择使用“any”或“all”“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) https://stackoverflow.com/questions/48428173/how-to-change-consecutive-repeating-values-in-pandas-dataframe-series-to-nan-or
# 2) https://stackoverflow.com/questions/19463985/pandas-drop-consecutive-duplicates
Input:
df should be a pandas.DataFrame of a a pandas.Series
Output:
df of ts with masked or droped values
'''
# Mask keeping the first occurence
if keep_first:
df = df.mask(df.shift(1) == df)
# Mask including the first occurence
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 occurence:")
print(Mask_Or_Drop_Successive_Identical_Values(ts, drop=False,
keep_first=True))
print("\n## 2) Mask including the first occurence:")
print(Mask_Or_Drop_Successive_Identical_Values(ts, drop=False,
keep_first=False))
print("\n## 3) Drop keeping the first occurence:")
print(Mask_Or_Drop_Successive_Identical_Values(ts, drop=True,
keep_first=True))
print("\n## 4) Drop including the first occurence:")
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 occurence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=False,
keep_first=True))
print("\n## 6) Mask including the first occurence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=False,
keep_first=False))
print("\n## 7) Drop 'any' keeping the first occurence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=True,
keep_first=True,
how='any'))
print("\n## 8) Drop 'all' keeping the first occurence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=True,
keep_first=True,
how='all'))
print("\n## 9) Drop 'any' including the first occurence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=True,
keep_first=False,
how='any'))
print("\n## 10) Drop 'all' including the first occurence:")
print(Mask_Or_Drop_Successive_Identical_Values(df, drop=True,
keep_first=False,
how='all'))
以下是预期结果：
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 occurence:
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 occurence:
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 occurence:
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 occurence:
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 occurence:
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 occurence:
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 occurence:
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 occurence:
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 occurence:
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 occurence:
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
只是另一种方式：
a.loc[a.ne(a.shift())]
方法pandas.Series.ne
是不相等运算符，因此a.ne（a.shift（））
相当于a！=a、 shift（）
。文档。这里有一个变体，它也将连续的NAN视为重复：
def删除连续的重复项：
#默认情况下，`shift`使用NaN作为填充值，这会中断
#删除连续的NAN。因此我们使用不同的哨兵
#反对。
shift=s.astype（object）.shift（-1，fill_value=object（））
返回s.loc[
（移位！=s）
&~（shift.isna（）&s.isna（））
]
创建新列
df['match'] = df.col1.eq(df.col1.shift())
然后：
我不明白为什么[147]和[175]的时间不同？你能解释一下你做了什么改变吗？因为我没有看到任何改变？也许是输入错误？@Biarys[175]
仅用于值的修改获取主要提升第一节，因此时间差。最初的一个在熊猫系列上工作，而修改后的一个在阵列上也在文章中列出。哦，我明白了。很难注意到从返回a[…]
到返回ar[…]
的变化。您的函数适用于数据帧吗？@Biarys对于数据帧，如果您要查找重复的行，我们只需使用切片：ar[：，：-1]！=ar[：，1::
，以及全部
减少。谢谢。我会尽量避免显式检查值如果keep_first:
足够（更好的样式）我们应该怎么做，如果我们想先做一个groupby，然后删除连续的重复项？例如df.groupby（['Col1'，'Col2']），并将其再次保存为数据帧？