调用resample后如何使用值0填充()？

Question

要么我不理解文档或它已过时.

如果我跑

user[["DOC_ACC_DT", "USER_SIGNON_ID"]].groupby("DOC_ACC_DT").agg(["count"]).resample("1D").fillna(value=0, method="ffill")

它得到了

TypeError: fillna() got an unexpected keyword argument 'value'

如果我跑

.fillna(0)

我明白了

ValueError: Invalid fill method. Expecting pad (ffill), backfill (bfill) or nearest. Got 0

如果我然后设置

.fillna(0, method="ffill") 

我明白了

TypeError: fillna() got multiple values for keyword argument 'method'

所以唯一有效的是

.fillna("ffill")

但当然这只是一个向前填补.但是,我想NaN用零替换.我在这做错了什么？

Answer 1

好吧,我不明白为什么上面的代码不能正常工作,我会等一个人给出一个比这个更好的答案,但我刚发现

.replace(np.nan, 0)

做我所期望的.fillna(0).

Answer 2

我做了一些测试，这很有趣。

样本：

import pandas as pd
import numpy as np

np.random.seed(1)
rng = pd.date_range('1/1/2012', periods=20, freq='S')
df = pd.DataFrame({'a':['a'] * 10 + ['b'] * 10,
                   'b':np.random.randint(0, 500, len(rng))}, index=rng)
df.b.iloc[3:8] = np.nan
print (df)
                     a      b
2012-01-01 00:00:00  a   37.0
2012-01-01 00:00:01  a  235.0
2012-01-01 00:00:02  a  396.0
2012-01-01 00:00:03  a    NaN
2012-01-01 00:00:04  a    NaN
2012-01-01 00:00:05  a    NaN
2012-01-01 00:00:06  a    NaN
2012-01-01 00:00:07  a    NaN
2012-01-01 00:00:08  a  335.0
2012-01-01 00:00:09  a  448.0
2012-01-01 00:00:10  b  144.0
2012-01-01 00:00:11  b  129.0
2012-01-01 00:00:12  b  460.0
2012-01-01 00:00:13  b   71.0
2012-01-01 00:00:14  b  237.0
2012-01-01 00:00:15  b  390.0
2012-01-01 00:00:16  b  281.0
2012-01-01 00:00:17  b  178.0
2012-01-01 00:00:18  b  276.0
2012-01-01 00:00:19  b  254.0

下采样：

可能的解决方案Resampler.asfreq：

如果使用asfreq，行为是相同的聚合first：

print (df.groupby('a').resample('2S').first())
                       a      b
a                              
a 2012-01-01 00:00:00  a   37.0
  2012-01-01 00:00:02  a  396.0
  2012-01-01 00:00:04  a    NaN
  2012-01-01 00:00:06  a    NaN
  2012-01-01 00:00:08  a  335.0
b 2012-01-01 00:00:10  b  144.0
  2012-01-01 00:00:12  b  460.0
  2012-01-01 00:00:14  b  237.0
  2012-01-01 00:00:16  b  281.0
  2012-01-01 00:00:18  b  276.0

print (df.groupby('a').resample('2S').first().fillna(0))
                       a      b
a                              
a 2012-01-01 00:00:00  a   37.0
  2012-01-01 00:00:02  a  396.0
  2012-01-01 00:00:04  a    0.0
  2012-01-01 00:00:06  a    0.0
  2012-01-01 00:00:08  a  335.0
b 2012-01-01 00:00:10  b  144.0
  2012-01-01 00:00:12  b  460.0
  2012-01-01 00:00:14  b  237.0
  2012-01-01 00:00:16  b  281.0
  2012-01-01 00:00:18  b  276.0

print (df.groupby('a').resample('2S').asfreq().fillna(0))
                       a      b
a                              
a 2012-01-01 00:00:00  a   37.0
  2012-01-01 00:00:02  a  396.0
  2012-01-01 00:00:04  a    0.0
  2012-01-01 00:00:06  a    0.0
  2012-01-01 00:00:08  a  335.0
b 2012-01-01 00:00:10  b  144.0
  2012-01-01 00:00:12  b  460.0
  2012-01-01 00:00:14  b  237.0
  2012-01-01 00:00:16  b  281.0
  2012-01-01 00:00:18  b  276.0

如果使用replace另一个值聚合为mean：

print (df.groupby('a').resample('2S').mean())
                           b
a                           
a 2012-01-01 00:00:00  136.0
  2012-01-01 00:00:02  396.0
  2012-01-01 00:00:04    NaN
  2012-01-01 00:00:06    NaN
  2012-01-01 00:00:08  391.5
b 2012-01-01 00:00:10  136.5
  2012-01-01 00:00:12  265.5
  2012-01-01 00:00:14  313.5
  2012-01-01 00:00:16  229.5
  2012-01-01 00:00:18  265.0

print (df.groupby('a').resample('2S').mean().fillna(0))
                           b
a                           
a 2012-01-01 00:00:00  136.0
  2012-01-01 00:00:02  396.0
  2012-01-01 00:00:04    0.0
  2012-01-01 00:00:06    0.0
  2012-01-01 00:00:08  391.5
b 2012-01-01 00:00:10  136.5
  2012-01-01 00:00:12  265.5
  2012-01-01 00:00:14  313.5
  2012-01-01 00:00:16  229.5
  2012-01-01 00:00:18  265.0

print (df.groupby('a').resample('2S').replace(np.nan,0))
                           b
a                           
a 2012-01-01 00:00:00  136.0
  2012-01-01 00:00:02  396.0
  2012-01-01 00:00:04    0.0
  2012-01-01 00:00:06    0.0
  2012-01-01 00:00:08  391.5
b 2012-01-01 00:00:10  136.5
  2012-01-01 00:00:12  265.5
  2012-01-01 00:00:14  313.5
  2012-01-01 00:00:16  229.5
  2012-01-01 00:00:18  265.0

上采样：

使用asfreq, 等同于replace：

print (df.groupby('a').resample('200L').asfreq().fillna(0))
                           a      b
a                                  
a 2012-01-01 00:00:00.000  a   37.0
  2012-01-01 00:00:00.200  0    0.0
  2012-01-01 00:00:00.400  0    0.0
  2012-01-01 00:00:00.600  0    0.0
  2012-01-01 00:00:00.800  0    0.0
  2012-01-01 00:00:01.000  a  235.0
  2012-01-01 00:00:01.200  0    0.0
  2012-01-01 00:00:01.400  0    0.0
  2012-01-01 00:00:01.600  0    0.0
  2012-01-01 00:00:01.800  0    0.0
  2012-01-01 00:00:02.000  a  396.0
  2012-01-01 00:00:02.200  0    0.0
  2012-01-01 00:00:02.400  0    0.0
  ...

print (df.groupby('a').resample('200L').replace(np.nan,0))
                               b
a                               
a 2012-01-01 00:00:00.000   37.0
  2012-01-01 00:00:00.200    0.0
  2012-01-01 00:00:00.400    0.0
  2012-01-01 00:00:00.600    0.0
  2012-01-01 00:00:00.800    0.0
  2012-01-01 00:00:01.000  235.0
  2012-01-01 00:00:01.200    0.0
  2012-01-01 00:00:01.400    0.0
  2012-01-01 00:00:01.600    0.0
  2012-01-01 00:00:01.800    0.0
  2012-01-01 00:00:02.000  396.0
  2012-01-01 00:00:02.200    0.0
  2012-01-01 00:00:02.400    0.0
  ...

print ((df.groupby('a').resample('200L').replace(np.nan,0).b == 
       df.groupby('a').resample('200L').asfreq().fillna(0).b).all())
True

结论：

对于下采样，使用相同的聚合函数，如sum, firstormean和 用于上采样asfreq。