4

你能解释一下这种奇怪的行为吗？

df=pd.DataFrame({'year':[1986,1987,1988],'bomb':arange(3)}).set_index('year')

In [9]: df.reindex(arange(1986,1988.125,.125))
Out[9]: 
          bomb
1986.000     0
1986.125   NaN
1986.250   NaN
1986.375   NaN
1986.500   NaN
1986.625   NaN
1986.750   NaN
1986.875   NaN
1987.000     1
1987.125   NaN
1987.250   NaN
1987.375   NaN
1987.500   NaN
1987.625   NaN
1987.750   NaN
1987.875   NaN
1988.000     2

In [10]: df.reindex(arange(1986,1988.1,.1))
Out[10]: 
        bomb
1986.0     0
1986.1   NaN
1986.2   NaN
1986.3   NaN
1986.4   NaN
1986.5   NaN
1986.6   NaN
1986.7   NaN
1986.8   NaN
1986.9   NaN
1987.0   NaN
1987.1   NaN
1987.2   NaN
1987.3   NaN
1987.4   NaN
1987.5   NaN
1987.6   NaN
1987.7   NaN
1987.8   NaN
1987.9   NaN
1988.0   NaN

当增量不是 .125 时，我发现新索引值不会“找到”具有匹配值的旧行。即有一个精度问题没有被克服。即使我在尝试插值之前强制索引为浮点数也是如此。发生了什么和/或这样做的正确方法是什么？通过使用，我已经能够让它以 0.1 的增量工作

reindex(  np.array(map(round,arange(1985,2010+dt,dt)*10))/10.0 )

顺便说一句，我这样做是线性插入多个列的第一步（例如“炸弹”就是其中之一）。如果有更好的方法可以做到这一点，我很乐意直截了当。

4

2 回答 2

0

You are getting what you ask for. The reindex method only tries to for the data onto the new index that you provide. As mentioned in comments you are probably looking for dates in the index. I guess you were expecting the reindex method to do this though(interpolation):

df2 =df.reindex(arange(1986,1988.125,.125))
pd.Series.interpolate(df2['bomb'])

1986.000    0.000
1986.125    0.125
1986.250    0.250
1986.375    0.375
1986.500    0.500
1986.625    0.625
1986.750    0.750
1986.875    0.875
1987.000    1.000
1987.125    1.125
1987.250    1.250
1987.375    1.375
1987.500    1.500
1987.625    1.625
1987.750    1.750
1987.875    1.875
1988.000    2.000
Name: bomb

the second example you use is inconsistency is probably because of floating point accuracies. Stepping by 0.125 is equal to 1/8 which can be exactly done in binary. stepping by 0.1 is not directly mappable to binary so 1987 is probably out by a fraction.

1987.0 == 1987.0000000001
False
于 2013-06-27T21:45:05.167 回答
0

我认为你最好使用 PeriodIndex 来做这样的事情

In [39]: df=pd.DataFrame({'bomb':np.arange(3)})

In [40]: df
Out[40]: 
   bomb
0     0
1     1
2     2

In [41]: df.index = pd.period_range('1986','1988',freq='Y').asfreq('M')

In [42]: df
Out[42]: 
         bomb
1986-12     0
1987-12     1
1988-12     2

In [43]: df = df.reindex(pd.period_range('1986','1988',freq='M'))

In [44]: df
Out[44]: 
         bomb
1986-01   NaN
1986-02   NaN
1986-03   NaN
1986-04   NaN
1986-05   NaN
1986-06   NaN
1986-07   NaN
1986-08   NaN
1986-09   NaN
1986-10   NaN
1986-11   NaN
1986-12     0
1987-01   NaN
1987-02   NaN
1987-03   NaN
1987-04   NaN
1987-05   NaN
1987-06   NaN
1987-07   NaN
1987-08   NaN
1987-09   NaN
1987-10   NaN
1987-11   NaN
1987-12     1
1988-01   NaN
In [45]: df.iloc[0,0] = -1

In [46]: df['interp'] = df['bomb'].interpolate()

In [47]: df
Out[47]: 
         bomb    interp
1986-01    -1 -1.000000
1986-02   NaN -0.909091
1986-03   NaN -0.818182
1986-04   NaN -0.727273
1986-05   NaN -0.636364
1986-06   NaN -0.545455
1986-07   NaN -0.454545
1986-08   NaN -0.363636
1986-09   NaN -0.272727
1986-10   NaN -0.181818
1986-11   NaN -0.090909
1986-12     0  0.000000
1987-01   NaN  0.083333
1987-02   NaN  0.166667
1987-03   NaN  0.250000
1987-04   NaN  0.333333
1987-05   NaN  0.416667
1987-06   NaN  0.500000
1987-07   NaN  0.583333
1987-08   NaN  0.666667
1987-09   NaN  0.750000
1987-10   NaN  0.833333
1987-11   NaN  0.916667
1987-12     1  1.000000
1988-01   NaN  1.000000
于 2013-06-28T00:31:39.033 回答