在熊猫中连接两行数据存储器

I need to concatenate two dataframes df_a anddf_b having equal number of rows (nRow) one after another without any consideration of keys. This function is similar to cbind in R programming language. The number of columns in each dataframe may be different.

我需要连接两个具有相等行数(nRow)的dataframes df_a和df_b，而不考虑键。该函数类似于R编程语言中的cbind。每个dataframe中的列数可能不同。

The resultant dataframe will have the same number of rows nRow and number of columns equal to the sum of number of columns in both the dataframes. In othe words, this is a blind columnar concatenation of two dataframes.

由此产生的dataframe将具有相同数量的行数和列数，等于dataframes中的列数的和。在othe中，这是两个dataframes的一个盲列连接。

import pandas as pd
dict_data = {'Treatment': ['C', 'C', 'C'], 'Biorep': ['A', 'A', 'A'], 'Techrep': [1, 1, 1], 'AAseq': ['ELVISLIVES', 'ELVISLIVES', 'ELVISLIVES'], 'mz':[500.0, 500.5, 501.0]}
df_a = pd.DataFrame(dict_data)
dict_data = {'Treatment1': ['C', 'C', 'C'], 'Biorep1': ['A', 'A', 'A'], 'Techrep1': [1, 1, 1], 'AAseq1': ['ELVISLIVES', 'ELVISLIVES', 'ELVISLIVES'], 'inte1':[1100.0, 1050.0, 1010.0]}
df_b = pd.DataFrame(dict_data)

2 个解决方案

#1

call concat and pass param axis=1 to concatenate column-wise:

调用concat并传递param轴=1以连接列:

In [5]:

pd.concat([df_a,df_b], axis=1)
Out[5]:
        AAseq Biorep  Techrep Treatment     mz      AAseq1 Biorep1  Techrep1  \
0  ELVISLIVES      A        1         C  500.0  ELVISLIVES       A         1   
1  ELVISLIVES      A        1         C  500.5  ELVISLIVES       A         1   
2  ELVISLIVES      A        1         C  501.0  ELVISLIVES       A         1   

  Treatment1  inte1  
0          C   1100  
1          C   1050  
2          C   1010

There is a useful guide to the various methods of merging, joining and concatenating online.

对于在线合并、连接和连接的各种方法，有一个有用的指南。

For example, as you have no *ing columns you can merge and use the indices as they have the same number of rows:

例如，由于没有冲突列，可以合并和使用索引，因为它们具有相同的行数:

In [6]:

df_a.merge(df_b, left_index=True, right_index=True)
Out[6]:
        AAseq Biorep  Techrep Treatment     mz      AAseq1 Biorep1  Techrep1  \
0  ELVISLIVES      A        1         C  500.0  ELVISLIVES       A         1   
1  ELVISLIVES      A        1         C  500.5  ELVISLIVES       A         1   
2  ELVISLIVES      A        1         C  501.0  ELVISLIVES       A         1   

  Treatment1  inte1  
0          C   1100  
1          C   1050  
2          C   1010

And for the same reasons as above a simple join works too:

基于同样的原因，一个简单的连接也可以工作:

In [7]:

df_a.join(df_b)
Out[7]:
        AAseq Biorep  Techrep Treatment     mz      AAseq1 Biorep1  Techrep1  \
0  ELVISLIVES      A        1         C  500.0  ELVISLIVES       A         1   
1  ELVISLIVES      A        1         C  500.5  ELVISLIVES       A         1   
2  ELVISLIVES      A        1         C  501.0  ELVISLIVES       A         1   

  Treatment1  inte1  
0          C   1100  
1          C   1050  
2          C   1010

#2

Thanks to @EdChum I was struggling with same problem especially when indexes do not match. Unfortunatly in pandas guide this case is missed (when you for example delete some rows)

多亏了@EdChum，我一直在努力解决同样的问题，尤其是当索引不匹配的时候。不幸的是，在熊猫指南中这个例子被忽略了(例如，当你删除一些行)

import pandas as pd
t=pd.DataFrame()
t['a']=[1,2,3,4]
t=t.loc[t['a']>1] #now index starts from 1

u=pd.DataFrame()
u['b']=[1,2,3] #index starts from 0

#option 1
#keep index of t
u.index = t.index 

#option 2
#index of t starts from 0
t.reset_index(drop=True, inplace=True)

#now concat will keep number of rows 
r=pd.concat([t,u], axis=1)

#1