Pandas Python:在一个数据框中合并每两行 [英] Pandas Python: Merging every two rows in one dataframe

问题描述

我如何获得

Idx            A B C
2004-04-01     1 1 0
2004-04-02     1 1 0
2004-05-01     0 0 0
2004-05-02     0 0 0

到

Idx            A B C
2004-04        2 2 0
2004-05        0 0 0

注意: 如何折叠索引(更具体地说，使索引仅转换为月份)和每两行折叠一次?

Notes: How do I collapse both the index (more specifically, making the index convert into just the month) and every two rows?

滚动是最好的方法吗?

更新-我简化了上面的版本，但是unutbu的答案似乎不起作用

UPDATE - I made the above version simple, but unutbu's answer does not seem to work

                       Time      A   B
1    2004-01-04 - 2004-01-10     0   0
2    2004-01-11 - 2004-01-17     0   0
3    2004-01-18 - 2004-01-24     0   0
4    2004-01-25 - 2004-01-31     0   0
5    2004-02-01 - 2004-02-07     56  0
6    2004-02-08 - 2004-02-14     67  0

推荐答案

您可以使用 groupby/sum操作:

You can aggregate rows using a groupby/sum operation:

import pandas as pd
import numpy as np

df = pd.DataFrame([('2004-04-01', 1L, 1L, 0L), ('2004-04-02', 1L, 1L, 0L),
       ('2004-05-01', 0L, 0L, 0L), ('2004-05-02', 0L, 0L, 0L)],
                  columns=['Idx', 'A', 'B', 'C'])
df['Idx'] = pd.DatetimeIndex(df['Idx'])

您可以按年份和月份分组:

You could group by the year and month:

print(df.groupby([d.strftime('%Y-%m') for d in df['Idx']]).sum())
#          A  B  C
# 2004-04  2  2  0
# 2004-05  0  0  0

# [2 rows x 3 columns]

或者，每两行分组一次:

Or, group by every two rows:

result = df.groupby(np.arange(len(df))//2).sum()
result.index = df.loc[1::2, 'Idx']
print(result)
#             A  B  C
# Idx                
# 2004-04-02  2  2  0
# 2004-05-02  0  0  0

# [2 rows x 3 columns]

注意:使用的是df.loc[1::2, 'Idx']而不是df.loc[::2, 'Idx']，因此汇总行的Idx对应于每个组中的第二个日期，而不是第一个日期.

Note: df.loc[1::2, 'Idx'] was used, instead of df.loc[::2, 'Idx'] so the Idx for the aggregated rows would correspond to the second date, not the first, in each group.

如果只需要年份和月份，则可以使用以下列表理解来设置索引:

If you want just the year and month, then you could use this list comprehension to set the index:

result.index = [d.strftime('%Y-%m') for d in df.loc[1::2, 'Idx']]
print(result)
#          A  B  C
# 2004-04  2  2  0
# 2004-05  0  0  0

# [2 rows x 3 columns]

但是，在处理日期时，使用DatetimeIndex作为索引而不是字符串值索引更为强大.因此，您可能希望保留DatetimeIndex，使用DatetimeIndex进行大部分工作，并仅在末尾将其转换为年月字符串以用于演示目的...

However, it's more powerful to have a DatetimeIndex for the index rather than a string-valued index when dealing with dates. So you might want to retain the DatetimeIndex, do most of your work with the DatetimeIndex, and just convert to a year-month string at the end for presentation purposes...

关于更新的问题:

import pandas as pd
import numpy as np

data = np.rec.array([('2004-01-04 - 2004-01-10', 0L, 0L),
       ('2004-01-11 - 2004-01-17', 0L, 0L),
       ('2004-01-18 - 2004-01-24', 0L, 0L),
       ('2004-01-25 - 2004-01-31', 0L, 0L),
       ('2004-02-01 - 2004-02-07', 56L, 0L),
       ('2004-02-08 - 2004-02-14', 67L, 0L)], 
      dtype=[('Time', 'O'), ('A', '<i8'), ('B', '<i8')])
df = pd.DataFrame(data)

具有一个包含两个日期的时间"列会使数据操作更加困难.最好有两个DatetimeIndex列，Start和End:

Having one Time column holding two dates makes data manipulation more difficult. It would be better to have two DatetimeIndex columns, Start and End:

df[['Start', 'End']] = df['Time'].str.extract('(?P<Start>.+) - (?P<End>.+)')
del df['Time']
df['Start'] = pd.DatetimeIndex(df['Start'])
df['End'] = pd.DatetimeIndex(df['End'])

然后您可以按Start列进行分组:

Then you could group by the Start column:

print(df.groupby([d.strftime('%Y-%m') for d in df['Start']]).sum())
#            A  B
# 2004-01    0  0
# 2004-02  123  0

# [2 rows x 2 columns]

或每两行分组一次，与以前基本相同:

Or group by every two rows, essentially the same as before:

result = df.groupby(np.arange(len(df))//2).sum()
result.index = df.loc[1::2, 'Start']
print(result)
#               A  B
# Start             
# 2004-01-11    0  0
# 2004-01-25    0  0
# 2004-02-08  123  0

# [3 rows x 2 columns]

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