Pandas合并/加入数据帧

Question

我不确定这意味着合并还是加入。在

我有一个包含以下列的dataframe ['times','spots']，dataframe a，我还有另一个dataframe，具有类似的列['times','spots']，dataframeb

我想更改数据帧A，使其具有来自数据帧B的值，这些值是从A中插值而来的。这样它将在数据帧A中有一个新的列spots_B

Answer 1

好吧，我要去一个壁架上，教你如何与后缀合并：

# First import our libraries
>>> import pandas as pd
>>> import numpy as np
# Then create our dataframes
>>> df_A = pd.DataFrame(np.random.rand(3,2),columns=['times','spots'])
>>> df_B = pd.DataFrame(np.random.rand(3,2),columns=['times','spots'])
# Set default values
>>> df_A['times'] = [1,2,3]
>>> df_B['times'] = [1,2,3]
>>> df_A['spots'] = [44,55,66]
>>> df_B['spots'] = [77,88,99]
# Here is what both dataframes contain
>>> df_A
   times  spots
0      1     44
1      2     55
2      3     66
>>> df_B
   times  spots
0      1     77
1      2     88
2      3     99
# Now the merge   note: this does not affect the first dataframe in place.
#                        It will create a new dataframe. You can overwrite the 
#                        first if you set the result to df_A instead of df_merged.
# Note the use of the keyword, suffixes. In the event that the same column names exist
#  in both dataframes (that aren't being merged on) Pandas will need to differentiate
#  between them. By default same column names will result in a '_x' will be appended to
#  the left dataframe column name, and a '_y' to the right dataframe column name
#  [order is set by the first two arguments in the merge function]. 
#  The suffixes keyword allows the user to override this behaviour with their
#  own version of '_x' and '_y'.
>>> df_merged = pd.merge(df_A,df_B,how='inner',on=['times'],suffixes=['_A','_B'])
>>> df_merged
   times  spots_A  spots_B
0      1       44       77
1      2       55       88
2      3       66       99

现在从您的问题看来，您不想修改spots的第一个dataframes列名。这可以用同样的方式实现，除了不使用suffixes=['_A','_B']使用suffixes=['','_B']。数据集的后缀基本上保持不变：

^{pr2}$

喂！我希望这有帮助。如果我误解了，而你实际上是在寻找A和B之间的插值值，告诉我，我会编辑这个答案。在

*编辑1*

考虑到你最后的评论，我相信你正在努力实现的目标。下面我将向您展示如何使用后缀扩展合并，然后使用“时间”插值方法用插值值填充spot_B中的nan

# Start by creating out datetimes to set for the times column
>>> times_A = []
>>> times_B = []
>>> for i in range(1,4):
...   times_A.append(datetime.datetime(year=2011,month=5,day=i))
...
>>> for i in range(1,6,2):
...   times_B.append(datetime.datetime(year=2011,month=5,day=i))
...
# times_A: May 1st, 2011 - May 3rd, 2011
>>> times_A
[datetime.datetime(2011, 5, 1, 0, 0), datetime.datetime(2011, 5, 2, 0, 0), datetime.datetime(2011, 5, 3, 0, 0)]
# times_B: May 1st 2011, May 3rd 2011, May 5th 2011
>>> times_B
[datetime.datetime(2011, 5, 1, 0, 0), datetime.datetime(2011, 5, 3, 0, 0), datetime.datetime(2011, 5, 5, 0, 0)]
# So now times_B is missing May 2nd, and has an extra time, May 5th.
>>> df_A['times'] = times_A
>>> df_B['times'] = times_B
>>> df_A['spots'] = [44,55,66]
>>> df_B['spots'] = [44,66,88]
>>> df_A
                times  spots
0 2011-05-01 00:00:00     44
1 2011-05-02 00:00:00     55
2 2011-05-03 00:00:00     66
>>> df_B
                times  spots
0 2011-05-01 00:00:00     44
1 2011-05-03 00:00:00     66
2 2011-05-05 00:00:00     88

# Now it appears you only care about the times in df_A - so
#   left merge df_A with df_B (include all times from df_A and  
#   try to merge with df_B or NaN). Below the date May 5th was dropped.
>>> df_merged = pd.merge(df_A,df_B,how='left',on=['times'],suffixes=['','_B'])
>>> df_merged
                times  spots  spots_B
0 2011-05-01 00:00:00     44       44
1 2011-05-02 00:00:00     55      NaN
2 2011-05-03 00:00:00     66       66

# Here is the important part:
# Since it appears that your data is going to be a time series
#   you will need to set your dataframe index to be the times column.
>>> df_merged = df_merged.set_index(['times'])
>>> df_merged
            spots  spots_B
times
2011-05-01     44       44
2011-05-02     55      NaN
2011-05-03     66       66

# With the times as index we can use the appropriate
#   interpolation method for best results
>>> df_merged['spots_B'] = df_merged['spots_B'].interpolate(method='time')
>>> df_merged
            spots  spots_B
times
2011-05-01     44       44
2011-05-02     55       55
2011-05-03     66       66

注意：在Series上，interpolate()的默认行为是假定每一行都是等距的。如果您的时间间隔不相等，则需要使用TimeSeries索引重新索引数据帧。当索引是timeseries时，可以在interpolate()函数中使用method='time'参数。

提问时尽量提供细节2。它有助于想回答问题的人完全理解您的问题。