2013-12-10 49 views
143

我有一個數據框從中刪除一些行。因此,我得到一個數據框,其索引是這樣的:[1,5,6,10,11],我想將它重置爲[0,1,2,3,4]。我該怎麼做?如何重置熊貓數據框中的索引?

ADDED

以下似乎工作:

df = df.reset_index() 
del df['index'] 

下不起作用:

df = df.reindex() 

回答

318

reset_index()是你在找什麼。如果你不希望它保存爲一列這樣做,那麼做:

df = df.reset_index(drop=True) 
+47

+1 'drop = True' – Rhubarb

+53

而不是重新分配數據幀到同一個變量,你可以設置'inplace = True'參數。 – ahuelamo

+1

請注意,如果'inplace = True',則方法返回無 – alyaxey

8

另一種解決方案是分配RangeIndexrange

df.index = pd.RangeIndex(len(df.index)) 

df.index = range(len(df.index)) 

這是更快:

df = pd.DataFrame({'a':[8,7], 'c':[2,4]}, index=[7,8]) 
df = pd.concat([df]*10000) 
print (df.head()) 

In [298]: %timeit df1 = df.reset_index(drop=True) 
The slowest run took 7.26 times longer than the fastest. This could mean that an intermediate result is being cached. 
10000 loops, best of 3: 105 µs per loop 

In [299]: %timeit df.index = pd.RangeIndex(len(df.index)) 
The slowest run took 15.05 times longer than the fastest. This could mean that an intermediate result is being cached. 
100000 loops, best of 3: 7.84 µs per loop 

In [300]: %timeit df.index = range(len(df.index)) 
The slowest run took 7.10 times longer than the fastest. This could mean that an intermediate result is being cached. 
100000 loops, best of 3: 14.2 µs per loop 
+0

@Outcast來源 - 最快的是'len(df.index)',381ns vs'df.shape' 1.17us。 Oyr有什麼缺失? – jezrael