Question

我有两个数据框df和times，分别代表维护记录和每月时间。我想根据times中的数据向df追加一列：

#df represents car maintenance records
data = {"07-18-2012": ["replaced wheels", 45, 200], "09-12-2014": ["changed oil", 30, 40], "09-18-2015": ["fixed dent", 92, 0]}
df = pd.DataFrame.from_dict(data, orient = "index")
df.index = pd.to_datetime(df.index)
df.sort_index(inplace = True)
df.columns = ["description", "mins_spent", "cost"]

#times represents monthly periods
rng = pd.date_range(start = '12/31/2013', end = '1/1/2015', freq='M')
ts = pd.Series(rng)
times = ts.to_frame(name = "months")

我正在尝试向days_since_maintenance添加名为times的新列，该列代表自df

以来最近一次维护以来的天数

我尝试使用df.ix[]，迭代for loop和searchsorted()。

df

                description  mins_spent  cost
2012-07-18  replaced wheels          45   200
2014-09-12      changed oil          30    40
2015-09-18       fixed dent          92     0

times

   months
0  2013-12-31
1  2014-01-31
2  2014-02-28
3  2014-03-31
4  2014-04-30
5  2014-05-31
6  2014-06-30
7  2014-07-31
8  2014-08-31
9  2014-09-30
10 2014-10-31
11 2014-11-30
12 2014-12-31

所需的DataFrame：

   months       days_since_maintenance
0  2013-12-31   531 days
1  2014-01-31   562 days
2  2014-02-28   ...
3  2014-03-31   ...
4  2014-04-30   ...
5  2014-05-31   ...
6  2014-06-30   ...
7  2014-07-31   ...
8  2014-08-31   774 days
9  2014-09-30   18 days
10 2014-10-31   ...
11 2014-11-30   ...
12 2014-12-31   ...

Answer 1

apply有效，但使用index而不是为dates添加新列：

def days_since_x(row, df):
    '''returns days between the row date
     and the most recent maintenance date in df'''

    #filter records
    all_maint_prior = df[(df.index <= row)]

    if all_maint_prior.empty:
        return float('NaN')

    else:
        #get last row of filtered results
        most_recent = all_maint_prior.iloc[-1]

        #return difference in dates
        return row-most_recent.name

times["days_since_maintenance"] = times["months"].apply(lambda row: days_since_x (row,df))

Answer 2

嗯，这绝对不是最好的解决方案，因为它会遍历您的df.index：

for d in df.index:
    times.ix[times['months'] >= d, 'days_since_maintenance'] = times['months'] - d

In [123]: times
Out[123]:
       months  days_since_maintenance
0  2013-12-31                531 days
1  2014-01-31                562 days
2  2014-02-28                590 days
3  2014-03-31                621 days
4  2014-04-30                651 days
5  2014-05-31                682 days
6  2014-06-30                712 days
7  2014-07-31                743 days
8  2014-08-31                774 days
9  2014-09-30                 18 days
10 2014-10-31                 49 days
11 2014-11-30                 79 days
12 2014-12-31                110 days

Answer 3

df['dates'] = df.index

def days_from_closest(x, df):
    closest = df[df['dates'] < x].ix[-1]
    return x - closest.dates

times['days_since_maintenance'] = times['months'].apply(lambda x: days_from_closest(x, df))

       months  days_since_maintenance
0  2013-12-31                531 days
1  2014-01-31                562 days
2  2014-02-28                590 days
3  2014-03-31                621 days
4  2014-04-30                651 days
5  2014-05-31                682 days
6  2014-06-30                712 days
7  2014-07-31                743 days
8  2014-08-31                774 days
9  2014-09-30                 18 days
10 2014-10-31                 49 days
11 2014-11-30                 79 days
12 2014-12-31                110 days

[13行x 2列]

根据另一个数据帧的内容向pandas数据帧添加一列

3 个答案: