比较Pandas DataFrame中的先前行值

Question

import pandas as pd
data={'col1':[1,3,3,1,2,3,2,2]}
df=pd.DataFrame(data,columns=['col1'])
print df


         col1  
    0     1          
    1     3          
    2     3          
    3     1          
    4     2          
    5     3          
    6     2          
    7     2          

我有以下Pandas DataFrame,我想创建另一个列,比较前一行col1,看看它们是否相等.最好的方法是什么？它就像下面的DataFrame.谢谢

    col1  match  
0     1   False     
1     3   False     
2     3   True     
3     1   False     
4     2   False     
5     3   False     
6     2   False     
7     2   True     

Answer 1

你需要eq有shift:

df['match'] = df.col1.eq(df.col1.shift())
print (df)
   col1  match
0     1  False
1     3  False
2     3   True
3     1  False
4     2  False
5     3  False
6     2  False
7     2   True

或者改为eq使用==,但在大型DataFrame中它有点慢:

df['match'] = df.col1 == df.col1.shift()
print (df)
   col1  match
0     1  False
1     3  False
2     3   True
3     1  False
4     2  False
5     3  False
6     2  False
7     2   True

时间:

import pandas as pd
data={'col1':[1,3,3,1,2,3,2,2]}
df=pd.DataFrame(data,columns=['col1'])
print (df)
#[80000 rows x 1 columns]
df = pd.concat([df]*10000).reset_index(drop=True)

df['match'] = df.col1 == df.col1.shift()
df['match1'] = df.col1.eq(df.col1.shift())
print (df)

In [208]: %timeit df.col1.eq(df.col1.shift())
The slowest run took 4.83 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 933 µs per loop

In [209]: %timeit df.col1 == df.col1.shift()
1000 loops, best of 3: 1 ms per loop

Answer 2

1）熊猫方法：使用diff：

df['match'] = df['col1'].diff().eq(0)

2) numpy 方法：使用np.ediff1d.

df['match'] = np.ediff1d(df['col1'].values, to_begin=np.NaN) == 0

两者都产生：

时间：（与DF@jezrael 使用的相同）

%timeit df.col1.eq(df.col1.shift())
1000 loops, best of 3: 731 µs per loop

%timeit df['col1'].diff().eq(0)
1000 loops, best of 3: 405 µs per loop

Answer 3

我很惊讶这里没有人提到滚动方法。滚动可轻松用于验证前 n 个值是否全部相同或执行任何自定义操作。这当然不如使用 diff 或 shift 快，但它可以轻松适应更大的窗口：

df['match'] = df['col1'].rolling(2).apply(lambda x: len(set(x)) != len(x),raw= True).replace({0 : False, 1: True})

Answer 4

这是一种基于NumPy数组的方法slicing，可让我们将视图用于输入数组以提高效率-

def comp_prev(a):
    return np.concatenate(([False],a[1:] == a[:-1]))

df['match'] = comp_prev(df.col1.values)

样品运行-

In [48]: df['match'] = comp_prev(df.col1.values)

In [49]: df
Out[49]: 
   col1  match
0     1  False
1     3  False
2     3   True
3     1  False
4     2  False
5     3  False
6     2  False
7     2   True

运行时测试-

In [56]: data={'col1':[1,3,3,1,2,3,2,2]}
    ...: df0=pd.DataFrame(data,columns=['col1'])
    ...: 

#@jezrael's soln1
In [57]: df = pd.concat([df0]*10000).reset_index(drop=True)

In [58]: %timeit df['match'] = df.col1 == df.col1.shift() 
1000 loops, best of 3: 1.53 ms per loop

#@jezrael's soln2
In [59]: df = pd.concat([df0]*10000).reset_index(drop=True)

In [60]: %timeit df['match'] = df.col1.eq(df.col1.shift())
1000 loops, best of 3: 1.49 ms per loop

#@Nickil Maveli's soln1   
In [61]: df = pd.concat([df0]*10000).reset_index(drop=True)

In [64]: %timeit df['match'] = df['col1'].diff().eq(0) 
1000 loops, best of 3: 1.02 ms per loop

#@Nickil Maveli's soln2
In [65]: df = pd.concat([df0]*10000).reset_index(drop=True)

In [66]: %timeit df['match'] = np.ediff1d(df['col1'].values, to_begin=np.NaN) == 0
1000 loops, best of 3: 1.52 ms per loop

# Posted approach in this post
In [67]: df = pd.concat([df0]*10000).reset_index(drop=True)

In [68]: %timeit df['match'] = comp_prev(df.col1.values)
1000 loops, best of 3: 376 µs per loop