pandas 数据帧 returns NaN 之间的计算

Question

我有一个名为 df_mod 的 pandas 数据框。此数据框中感兴趣的一个变量称为 Evap_mod。当我使用命令 print(df_mod['Evap_mod']) 时，它 returns:

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

我有另一个名为 dff 的 pandas 数据框。此数据框中感兴趣的一个变量称为 PET_PT。当我使用命令 print(dff['PET_PT']) 时，它 returns:

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

我想运行这两个变量之间的简单计算：

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，这个计算只是 returns 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

有谁知道为什么它 returns NaN 以及如何解决这个问题？

Answer 1

原因是索引值不同，所以除法后索引值不匹配并创建了 NaNs。

解决方案是 map 系列 ddf['PET_PT'] 由 DatetimeIndex.normalize 创建的辅助列 date 用于删除时间并使用 pandas means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

如果长度相同DataFrame且索引值仅相差倍数，您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

您的列包含缺失数据，因此您应该根据您的 objective

使用不同的方法（均值、零、中值、随机等）来估算值

Answer 3

pandas 和 numpy 行为之间存在差异。每当您计算 np.mean(x) 如果 x 包含 NaN 时，您将得到 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。

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