python – 如何在不复制数据的情况下连接pandas DataFrame?

我想连接两个pandas DataFrames而不复制数据.也就是说,我希望连接的DataFrame是两个原始DataFrame中数据的视图.我尝试使用concat(),但没有用.此代码块显示更改基础数据会影响连接的两个DataFrame,但不会影响连接的DataFrame:

arr = np.random.randn(12).reshape(6, 2)
df = pd.DataFrame(arr, columns = ('VALE5', 'PETR4'), index = dates)
arr2 = np.random.randn(12).reshape(6, 2)
df2 = pd.DataFrame(arr, columns = ('AMBV3', 'BBDC4'), index = dates)
df_concat = pd.concat(dict(A = df, B = df2),axis=1)
pp(df)
pp(df_concat)
arr[0, 0] = 9999999.99
pp(df)
pp(df_concat)

这是最后五行的输出. df在为arr [0,0]分配新值后更改; df_concat没有受到影响.

In [56]: pp(df)
           VALE5     PETR4
2013-01-01 -0.557180  0.170073
2013-01-02 -0.975797  0.763136
2013-01-03 -0.913254  1.042521
2013-01-04 -1.973013 -2.069460
2013-01-05 -1.259005  1.448442
2013-01-06 -0.323640  0.024857

In [57]: pp(df_concat)
               A                   B          
           VALE5     PETR4     AMBV3     BBDC4
2013-01-01 -0.557180  0.170073 -0.557180  0.170073
2013-01-02 -0.975797  0.763136 -0.975797  0.763136
2013-01-03 -0.913254  1.042521 -0.913254  1.042521
2013-01-04 -1.973013 -2.069460 -1.973013 -2.069460
2013-01-05 -1.259005  1.448442 -1.259005  1.448442
2013-01-06 -0.323640  0.024857 -0.323640  0.024857

In [58]: arr[0, 0] = 9999999.99

In [59]: pp(df)
                 VALE5     PETR4
2013-01-01  9999999.990000  0.170073
2013-01-02       -0.975797  0.763136
2013-01-03       -0.913254  1.042521
2013-01-04       -1.973013 -2.069460
2013-01-05       -1.259005  1.448442
2013-01-06       -0.323640  0.024857

In [60]: pp(df_concat)
               A                   B          
           VALE5     PETR4     AMBV3     BBDC4
2013-01-01 -0.557180  0.170073 -0.557180  0.170073
2013-01-02 -0.975797  0.763136 -0.975797  0.763136
2013-01-03 -0.913254  1.042521 -0.913254  1.042521
2013-01-04 -1.973013 -2.069460 -1.973013 -2.069460
2013-01-05 -1.259005  1.448442 -1.259005  1.448442
2013-01-06 -0.323640  0.024857 -0.323640  0.024857

我想这意味着concat()创建了一个数据副本.有没有办法避免复制? (我想尽量减少内存使用量).

此外,还有一种快速方法可以检查两个DataFrame是否链接到相同的基础数据? (除了经历更改数据和检查每个DataFrame是否已更改)的麻烦之外

谢谢您的帮助.

FS

最佳答案 你不能(至少很容易).当你调用concat时,最终会调用np.concatenate.

请参阅this answer explaining why you can’t concatenate arrays without copying.缺点是数组不保证在内存中是连续的.

这是一个简单的例子

a = rand(2, 10)
x, y = a
z = vstack((x, y))
print 'x.base is a and y.base is a ==', x.base is a and y.base is a
print 'x.base is z or y.base is z ==', x.base is z or y.base is z

输出:

x.base is a and y.base is a == True
x.base is z or y.base is z == False

即使x和y共享相同的基数,即a,连接(因此vstack)也不能假设它们确实存在,因为人们经常想要连接任意跨越的数组.

您可以轻松生成两个具有不同步长的数组,共享相同的内存,如下所示:

a = arange(10)
b = a[::2]
print a.strides
print b.strides

输出:

(8,)
(16,)

这就是为什么会发生以下情况:

In [214]: a = arange(10)

In [215]: a[::2].view(int16)
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-215-0366fadb1128> in <module>()
----> 1 a[::2].view(int16)

ValueError: new type not compatible with array.

In [216]: a[::2].copy().view(int16)
Out[216]: array([0, 0, 0, 0, 2, 0, 0, 0, 4, 0, 0, 0, 6, 0, 0, 0, 8, 0, 0, 0], dtype=int16)

编辑:当df1.dtype!= df2.dtype不会复制时,使用pd.merge(df1,df2,copy = False)(或df1.merge(df2,copy = False)).否则,制作副本.

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