pandas 大熊猫系列上的 numpy 差异
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numpy diff on a pandas Series
提问by Dan Allan
I want to use numpy.diff on a pandas Series. Am I right that this is a bug? Or am I doing it wrong?
我想在Pandas系列上使用 numpy.diff。我说这是一个错误是正确的吗？还是我做错了？
In [163]: s = Series(np.arange(10))
In [164]: np.diff(s)
Out[164]:
0 NaN
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 NaN
In [165]: np.diff(np.arange(10))
Out[165]: array([1, 1, 1, 1, 1, 1, 1, 1, 1])
I am using pandas 0.9.1rc1, numpy 1.6.1.
我正在使用Pandas 0.9.1rc1，numpy 1.6.1。
回答by Aman
Pandas implements difflike so:
Pandasdiff是这样实现的：
In [3]: s = pd.Series(np.arange(10))
In [4]: s.diff()
Out[4]:
0 NaN
1 1
2 1
3 1
4 1
5 1
6 1
7 1
8 1
9 1
Using np.diffdirectly:
np.diff直接使用：
In [7]: np.diff(s.values)
Out[7]: array([1, 1, 1, 1, 1, 1, 1, 1, 1])
In [8]: np.diff(np.array(s))
Out[8]: array([1, 1, 1, 1, 1, 1, 1, 1, 1])
So why doesn't np.diff(s)work? Because np is taking np.asanyarray()of the series before finding the diff. Like so:
那么为什么不起作用np.diff(s)呢？因为 npnp.asanyarray()在找到diff. 像这样：
In [25]: a = np.asanyarray(s)
In [26]: a
Out[26]:
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
In [27]: np.diff(a)
Out[27]:
0 NaN
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 NaN
