扩展数据框中的每一行

debugcn 发表于 Dev

ℕʘʘḆḽḘ

考虑这个简单的例子

data = pd.DataFrame({'mydate' : [pd.to_datetime('2016-06-06'),
                                 pd.to_datetime('2016-06-02')],
                     'value' : [1, 2]})

data.set_index('mydate', inplace = True)

data
Out[260]: 
            value
mydate           
2016-06-06      1
2016-06-02      2

我想遍历每一行，以便数据框围绕当前行的每个索引值（即日期）“放大”几天（前 2 天，后 2 天）。

举例来说，如果你考虑的第一行，我想告诉熊猫加4个排，对应天2016-06-04，2016-06-05，2016-06-07和2016-06-07。将value这些多余的行可以只是whathever是value该行（在这种情况下：1）。该逻辑适用于每一行，最终的数据帧是所有这些放大的数据帧的串联。

我在一个中尝试了以下功能apply(., axis = 1)：

def expand_onerow(df, ndaysback = 2, nhdaysfwd = 2):

    new_index = pd.date_range(pd.to_datetime(df.name) - pd.Timedelta(days=ndaysback), 
                              pd.to_datetime(df.name) + pd.Timedelta(days=nhdaysfwd), 
                              freq='D')

    newdf = df.reindex(index=new_index, method='nearest')     #New df with expanded index
    return newdf

但不幸的是，我跑步data.apply(lambda x: expand_onerow(x), axis = 1)给出了：

  File "pandas/_libs/tslib.pyx", line 1165, in pandas._libs.tslib._Timestamp.__richcmp__

TypeError: ("Cannot compare type 'Timestamp' with type 'str'", 'occurred at index 2016-06-06 00:00:00')

我尝试的另一种方法如下：我首先重置索引，

data.reset_index(inplace = True)
data
Out[339]: 
      mydate  value
0 2016-06-06      1
1 2016-06-02      2

然后我稍微修改一下我的功能

def expand_onerow_alt(df, ndaysback = 2, nhdaysfwd = 2):

    new_index = pd.date_range(pd.to_datetime(df.mydate) - pd.Timedelta(days=ndaysback), 
                              pd.to_datetime(df.mydate) + pd.Timedelta(days=nhdaysfwd), 
                              freq='D')
    newdf = pd.Series(df).reindex(index = new_index).T  #New df with expanded index
    return newdf

这使

data.apply(lambda x: expand_onerow_alt(x), axis = 1)
Out[338]: 
   2016-05-31  2016-06-01  2016-06-02  2016-06-03  2016-06-04  2016-06-05  2016-06-06  2016-06-07  2016-06-08
0         nan         nan         nan         nan         nan         nan         nan         nan         nan
1         nan         nan         nan         nan         nan         nan         nan         nan         nan

更近了，但还没有...

我不明白这里有什么问题。我错过了什么？我在这里寻找最潘多尼的方法。

谢谢！

贝尼

我修改了一点你的功能

def expand_onerow(df, ndaysback = 2, nhdaysfwd = 2):

    new_index = pd.date_range(pd.to_datetime(df.index[0]) - pd.Timedelta(days=ndaysback),
                              pd.to_datetime(df.index[0]) + pd.Timedelta(days=nhdaysfwd),
                              freq='D')

    newdf = df.reindex(index=new_index, method='nearest')     #New df with expanded index
    return newdf

pd.concat([expand_onerow(data.loc[[x],:], ndaysback = 2, nhdaysfwd = 2) for x ,_ in data.iterrows()])


Out[455]: 
            value
2016-05-31      2
2016-06-01      2
2016-06-02      2
2016-06-03      2
2016-06-04      2
2016-06-04      1
2016-06-05      1
2016-06-06      1
2016-06-07      1
2016-06-08      1

更多信息

基本上那一行等于

l=[]
for x ,_ in data.iterrows():

    l.append(expand_onerow(data.loc[[x],:], ndaysback = 2, nhdaysfwd = 2))# query out each row by using their index(x is the index for each row) and append then into a empty list


pd.concat(l)# concat the list to one df at the end

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