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我想我遇到了熊猫的一个错误。我希望得到一些帮助来验证错误或帮助我找出我的逻辑错误在我的代码中的位置。

我的代码如下：

import pandas, numpy, StringIO

def sq_fixer(sr):
    sr = sr.where(sr != '20200229')
    ranks = sr.argsort().astype(float)
    ranks[ranks == -1] = numpy.nan

    return ','.join(ranks.astype(numpy.str))

def correct_date(sr):

    date_fixer = lambda x: pandas.datetime(x.year -100, x.month, x.day) if x > pandas.datetime.now() else x
    sr = pandas.to_datetime(sr).apply(date_fixer).astype(pandas.datetime)

    return sr 

txt = '''ID,RUN_START_DATE,PUSHUP_START_DATE,SITUP_START_DATE,PULLUP_START_DATE
1,2013-01-24,2013-01-02,,2013-02-03
2,2013-01-30,2013-01-21,2013-01-13,2013-01-06
3,2013-01-29,2013-01-28,2013-01-01,2013-01-29
4,2013-02-16,2013-02-12,2013-01-04,2013-02-11
5,2013-01-06,2013-02-07,2013-02-25,2013-02-12
6,2013-01-26,2013-01-28,2013-02-12,2013-01-10
7,2013-01-26,,2013-01-12,2013-01-30
8,2013-01-03,2013-01-24,2013-01-19,2013-01-02
9,2013-01-22,2013-01-13,2013-02-03,
10,2013-02-06,2013-01-16,2013-02-07,2013-01-11
3347,,2008-02-27,2008-04-10,2008-02-13 
3588,2004-09-12,,2004-11-06,2004-09-06 
3784,2003-02-22,,2003-06-21,2003-02-19 
593,2009-04-03,,2009-06-01,2009-04-01 
4148,2003-03-21,2002-09-20,2003-04-01,2003-01-01 
4299,2004-05-24,2004-07-23,,2004-04-22 
4590,2005-05-05,2005-12-05,2005-04-05,
4830,2001-06-12,2000-10-12,2001-07-28,2001-01-28 
4941,2006-11-08,2006-12-19,2006-07-19,2007-02-24 
1416,2004-04-03,2004-05-19,2004-02-06,
1580,2008-12-20,,2009-03-19,2008-12-19 
1661,2005-10-03,2005-10-26,2005-09-12,2006-02-19 
1759,2001-10-18,,2002-01-17,2001-10-17 
1858,2003-04-14,2003-05-17,,2002-12-17 
1972,2003-06-01,2003-07-14,2002-12-14,
5905,2000-11-18,2001-01-13,,2000-11-04 
2052,2002-06-11,,2002-08-23,2001-12-12 
2165,2006-10-01,,2007-02-27,2006-09-30 
2218,2007-09-19,,2008-02-06,2007-09-09 
2350,2000-08-08,,2000-09-22,2000-01-08 
2432,2001-08-22,,2001-09-25,2000-12-16 
2611,2005-05-07,,2005-06-05,2005-03-26 
2612,2005-05-06,,2005-05-26,2005-04-11 
7378,2009-08-07,2009-01-30,2010-01-20,2009-06-08 
7550,2006-04-08,,2006-06-01,2006-04-01  '''

df = pandas.read_csv(StringIO.StringIO(txt))

sequence_array = ['RUN_START_DATE', 'PUSHUP_START_DATE', 'SITUP_START_DATE', 'PULLUP_START_DATE']
xsequence_array = ['X_RUN_START_DATE', 'X_PUSHUP_START_DATE', 'X_SITUP_START_DATE', 'X_PULLUP_START_DATE']

df[sequence_array] = df[sequence_array].apply(correct_date, axis=1)

fix_day = lambda x: x if x > 0 else 29
fix_month = lambda x: x if x > 0 else 02
fix_year = lambda x: x if x > 0 else 2020

for col in sequence_array:

    xcol = 'X_{0}'.format(col)
    df[xcol] = ['{0:04d}{1:02d}{2:02d}'.format(fix_year(c.year), fix_month(c.month), fix_day(c.day)) for c in df[col]]

df['X_AS_SEQUENCE'] = df[xsequence_array].apply(sq_fixer, axis=1)

当我运行代码时，大多数结果都是正确的。以索引 6 为例：

In [31]: df.ix[6]
Out[31]: 
ID                                       7
RUN_START_DATE         2013-01-26 00:00:00
PUSHUP_START_DATE                      NaN
SITUP_START_DATE       2013-01-12 00:00:00
PULLUP_START_DATE      2013-01-30 00:00:00
X_RUN_START_DATE                  20130126
X_PUSHUP_START_DATE               20200229
X_SITUP_START_DATE                20130112
X_PULLUP_START_DATE               20130130
X_AS_SEQUENCE              1.0,nan,0.0,2.0

但是，某些索引似乎会抛出 pandas.argsort() 循环。以索引 10 为例：

In [32]: df.ix[10]
Out[32]: 
ID                                    3347
RUN_START_DATE                         NaN
PUSHUP_START_DATE      2008-02-27 00:00:00
SITUP_START_DATE       2008-04-10 00:00:00
PULLUP_START_DATE      2008-02-13 00:00:00
X_RUN_START_DATE                  20200229
X_PUSHUP_START_DATE               20080227
X_SITUP_START_DATE                20080410
X_PULLUP_START_DATE               20080213
X_AS_SEQUENCE              nan,2.0,0.0,1.0

argsort 应该返回nan,1.0,2.0,0.0而不是nan,2.0,0.0,1.0.

我已经坚持了三天。在这一点上，我不确定是我还是错误。我不知道如何回溯它以获得答案。非常感激任何的帮助！

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2 回答 2

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您可能对结果的解释argsort不正确。 argsort没有给出值的排名。如果要对值进行排名，请使用排名方法。

返回的 Series 中的值argsort给出了删除 NaN 后原始值的对应位置。在您的情况下，由于您将 20200229 转换为 NaN，因此您是 argsorting NaN, 20080227, 20080410, 20080213。非 NaN 值是

nonnan = [20080227, 20080410, 20080213]

结果，NaN, 2, 0, 1说：

argsort     sorted values
  NaN       NaN
   2        nonnan[2] = 20080213
   0        nonnan[0] = 20080227
   1        nonnan[1] = 20080410

所以在我看来还可以。

于 2013-03-26T07:05:14.337 回答
0

如果要对系列进行排序，只需使用 sort_values() 或 rank() 函数：

In [2]: a=pd.Series([3,2,1])

In [3]: a
Out[3]:
0    3
1    2
2    1
dtype: int64
In [4]: a.sort_values()
Out[4]:
2    1
1    2
0    3
dtype: int64

如果您使用 argsort()，这将为您提供已排序系列中每个元素的位置，在这种情况下，1 应位于 0 位置，2 应位于 1 位置，3 应位于 2 位置

In [5]: a.argsort()
Out[5]:
0    2
1    1
2    0
dtype: int64
于 2016-03-28T16:13:23.840 回答