pandas dataframe之间的计算返回NaN

Question

我有一个名为df_mod的pandas数据帧。 此数据Evap_mod一个感兴趣的变量称为Evap_mod 。 当我使用命令print(df_mod['Evap_mod']) ，它返回：

2003-12-20 00:30:00    1.930664
2003-12-21 00:30:00    1.789290
2003-12-22 00:30:00    2.318347
2003-12-23 00:30:00    1.741943
2003-12-24 00:30:00    1.686124
2003-12-25 00:30:00    1.852876
2003-12-26 00:30:00    1.759650
2003-12-27 00:30:00    1.566521
2003-12-28 00:30:00    1.496039
2003-12-29 00:30:00    1.540751
2003-12-30 00:30:00    2.006475
2003-12-31 00:30:00    1.920912
Name: Evap_mod, Length: 729, dtype: float32

我有另一个叫做dff pandas数据帧。 该数据帧中的一个感兴趣的变量称为PET_PT 。 当我使用命令print(dff['PET_PT']) ，它返回：

2003-12-20    4.810697
2003-12-21    4.739378
2003-12-22    4.994467
2003-12-23    5.138086
2003-12-24    5.024226
2003-12-25    4.937206
2003-12-26    4.551416
2003-12-27         NaN
2003-12-28         NaN
2003-12-29         NaN
2003-12-30         NaN
2003-12-31         NaN
Freq: D, Name: PET_PT, Length: 729, dtype: float64

我想在这两个变量之间运行简单的以下计算：

df_mod['ER_mod']=(df_mod['Evap_mod']+np.mean(ddf['PET_PT']))/(ddf['PET_PT']+np.mean(ddf['PET_PT']))

不幸的是，这个计算只返回NaN：

2003-12-20 00:30:00   NaN
2003-12-21 00:30:00   NaN
2003-12-22 00:30:00   NaN
2003-12-23 00:30:00   NaN
2003-12-24 00:30:00   NaN
2003-12-25 00:30:00   NaN
2003-12-26 00:30:00   NaN
2003-12-27 00:30:00   NaN
2003-12-28 00:30:00   NaN
2003-12-29 00:30:00   NaN
2003-12-30 00:30:00   NaN
2003-12-31 00:30:00   NaN
Name: ER_mod, Length: 729, dtype: float64

有没有人知道为什么它返回NaN以及如何克服这个问题？

Answer 1

原因是不同的索引值，所以在除以索引值之后不匹配并创建NaN s。

解决方案是map Series ddf['PET_PT']由DatetimeIndex.normalize创建的辅助列date删除时间，并使用pandas mean s函数：

#same index values like df_mod
new = df_mod.assign(date = df_mod.index.normalize())['date'].map(ddf['PET_PT'])
print (new)
2003-12-20 00:30:00    4.810697
2003-12-21 00:30:00    4.739378
2003-12-22 00:30:00    4.994467
2003-12-23 00:30:00    5.138086
2003-12-24 00:30:00    5.024226
2003-12-25 00:30:00    4.937206
2003-12-26 00:30:00    4.551416
2003-12-27 00:30:00         NaN
2003-12-28 00:30:00         NaN
2003-12-29 00:30:00         NaN
2003-12-30 00:30:00         NaN
2003-12-31 00:30:00         NaN
Name: date, dtype: float64

df_mod['ER_mod']= df_mod['Evap_mod'] + ddf['PET_PT'].mean())/(new+ddf['PET_PT'].mean()
print (df_mod)
                     Evap_mod    ER_mod
2003-12-20 00:30:00  1.930664  0.702960
2003-12-21 00:30:00  1.789290  0.693480
2003-12-22 00:30:00  2.318347  0.729125
2003-12-23 00:30:00  1.741943  0.661170
2003-12-24 00:30:00  1.686124  0.663134
2003-12-25 00:30:00  1.852876  0.685986
2003-12-26 00:30:00  1.759650  0.704152
2003-12-27 00:30:00  1.566521       NaN
2003-12-28 00:30:00  1.496039       NaN
2003-12-29 00:30:00  1.540751       NaN
2003-12-30 00:30:00  2.006475       NaN
2003-12-31 00:30:00  1.920912       NaN

如果相同长度的DataFrame和inde值的差异是次数，则可以将一个索引重新分配给另一个索引：

ddf.index = df_mod.index

df_mod['ER_mod'] = (df_mod['Evap_mod'] + ddf['PET_PT'].mean())/\
                   (ddf['PET_PT'] + ddf['PET_PT'].mean())
print (df_mod)
                     Evap_mod    ER_mod
2003-12-20 00:30:00  1.930664  0.702960
2003-12-21 00:30:00  1.789290  0.693480
2003-12-22 00:30:00  2.318347  0.729125
2003-12-23 00:30:00  1.741943  0.661170
2003-12-24 00:30:00  1.686124  0.663134
2003-12-25 00:30:00  1.852876  0.685986
2003-12-26 00:30:00  1.759650  0.704152
2003-12-27 00:30:00  1.566521       NaN
2003-12-28 00:30:00  1.496039       NaN
2003-12-29 00:30:00  1.540751       NaN
2003-12-30 00:30:00  2.006475       NaN
2003-12-31 00:30:00  1.920912       NaN

Answer 2

您的列包含缺失的数据，因此您应该根据您的目标通过不同的方法（平均值，零，中位数，随机等）来估算值

Answer 3

这里有pandas和numpy行为之间的区别。 无论何时计算np.mean(x)如果x包含NaN您将使用NaN作为结果，同时使用pandas NaN将被忽略。 以下应该有效

df_mod['ER_mod'] = (df_mod['Evap_mod'] + ddf['PET_PT'].mean())/\
                   (ddf['PET_PT'] + ddf['PET_PT'].mean())

否则你可以使用np.nanmean而不是np.mean 。