从函数向Pandas Dataframe添加多个列

Question

我有一个mydf有两列的pandas数据框,两列都是datetime数据类型:mydate和mytime.我要添加三个多列:hour,weekday,和weeknum.

def getH(t): #gives the hour
    return t.hour
def getW(d): #gives the week number
    return d.isocalendar()[1] 
def getD(d): #gives the weekday
    return d.weekday() # 0 for Monday, 6 for Sunday

mydf["hour"] = mydf.apply(lambda row:getH(row["mytime"]), axis=1)
mydf["weekday"] = mydf.apply(lambda row:getD(row["mydate"]), axis=1)
mydf["weeknum"] = mydf.apply(lambda row:getW(row["mydate"]), axis=1)

该代码段有效,但它的计算效率不高,因为它至少循环数据帧三次.我想知道是否有更快和/或更优化的方法来做到这一点.例如,使用zip或merge？例如,如果我只创建一个返回三个元素的函数,我该如何实现呢？为了说明,该功能将是:

def getHWd(d,t):
    return t.hour, d.isocalendar()[1], d.weekday()

Answer 1

这是使用一个方法来做到这一点 apply

说,df就像

In [64]: df
Out[64]:
       mydate     mytime
0  2011-01-01 2011-11-14
1  2011-01-02 2011-11-15
2  2011-01-03 2011-11-16
3  2011-01-04 2011-11-17
4  2011-01-05 2011-11-18
5  2011-01-06 2011-11-19
6  2011-01-07 2011-11-20
7  2011-01-08 2011-11-21
8  2011-01-09 2011-11-22
9  2011-01-10 2011-11-23
10 2011-01-11 2011-11-24
11 2011-01-12 2011-11-25

为了便于阅读,我们将lambda函数分离出来,并将其定义为

In [65]: lambdafunc = lambda x: pd.Series([x['mytime'].hour,
                                           x['mydate'].isocalendar()[1],
                                           x['mydate'].weekday()])

并且,apply将结果存储到df[['hour', 'weekday', 'weeknum']]

In [66]: df[['hour', 'weekday', 'weeknum']] = df.apply(lambdafunc, axis=1)

而且,输出就像

In [67]: df
Out[67]:
       mydate     mytime  hour  weekday  weeknum
0  2011-01-01 2011-11-14     0       52        5
1  2011-01-02 2011-11-15     0       52        6
2  2011-01-03 2011-11-16     0        1        0
3  2011-01-04 2011-11-17     0        1        1
4  2011-01-05 2011-11-18     0        1        2
5  2011-01-06 2011-11-19     0        1        3
6  2011-01-07 2011-11-20     0        1        4
7  2011-01-08 2011-11-21     0        1        5
8  2011-01-09 2011-11-22     0        1        6
9  2011-01-10 2011-11-23     0        2        0
10 2011-01-11 2011-11-24     0        2        1
11 2011-01-12 2011-11-25     0        2        2

Answer 2

为了补充约翰高尔特的答案:

根据执行的任务lambdafunc,您可以通过将结果存储apply为新的DataFrame然后加入原始文件来体验一些加速:

lambdafunc = lambda x: pd.Series([x['mytime'].hour,
                                  x['mydate'].isocalendar()[1],
                                  x['mydate'].weekday()])

newcols = df.apply(lambdafunc, axis=1)
newcols.columns = ['hour', 'weekday', 'weeknum']
newdf = df.join(newcols) 

即使你没有看到提速,我也建议使用join.您可以避免SettingWithCopyWarning直接在列上分配时弹出的(总是很烦人):

SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

Answer 3

您可以通过让您应用的函数返回pd.Series带有命名元素的函数来以更简洁的方法执行此操作：

def process(row):
    return pd.Series(dict(b=row["a"] * 2, c=row["a"] + 2))


my_df = pd.DataFrame(dict(a=range(10)))
new_df = my_df.join(my_df.apply(process, axis="columns"))

结果是：

   a   b   c
0  0   0   2
1  1   2   3
2  2   4   4
3  3   6   5
4  4   8   6
5  5  10   7
6  6  12   8
7  7  14   9
8  8  16  10
9  9  18  11