在熊猫中连接两个数据框架的行

我需要连接两个数据帧 df_a和 df_b具有相同的行数(nRow)一个接一个没有任何关键考虑。这个功能类似于 R programming language中的 cbind。每个数据框中的列数可能不同。

结果数据框架将具有相同数量的行 nRow和相等于两个数据框架中列数之和的列数。换句话说，这是两个数据帧的盲柱状串联。

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)

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最佳答案

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

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.

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

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:

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

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)