下面是一个合并数据帧字典的方法，同时保持列名与字典同步。如果需要，它还会填充缺失的值:

这是合并数据帧字典的函数

def MergeDfDict(dfDict, onCols, how='outer', naFill=None):
  keys = dfDict.keys()
  for i in range(len(keys)):
    key = keys[i]
    df0 = dfDict[key]
    cols = list(df0.columns)
    valueCols = list(filter(lambda x: x not in (onCols), cols))
    df0 = df0[onCols + valueCols]
    df0.columns = onCols + [(s + '_' + key) for s in valueCols] 

    if (i == 0):
      outDf = df0
    else:
      outDf = pd.merge(outDf, df0, how=how, on=onCols)   

  if (naFill != None):
    outDf = outDf.fillna(naFill)

  return(outDf)

好的，让我们生成数据并测试:

def GenDf(size):
  df = pd.DataFrame({'categ1':np.random.choice(a=['a', 'b', 'c', 'd', 'e'], size=size, replace=True),
                      'categ2':np.random.choice(a=['A', 'B'], size=size, replace=True), 
                      'col1':np.random.uniform(low=0.0, high=100.0, size=size), 
                      'col2':np.random.uniform(low=0.0, high=100.0, size=size)
                      })
  df = df.sort_values(['categ2', 'categ1', 'col1', 'col2'])
  return(df)


size = 5
dfDict = {'US':GenDf(size), 'IN':GenDf(size), 'GER':GenDf(size)}   
MergeDfDict(dfDict=dfDict, onCols=['categ1', 'categ2'], how='outer', naFill=0)

在python 3.6.3和pandas 0.22.0中，你也可以使用concat，只要你将你想要用于连接的列设置为index:

pd.concat(
    objs=(iDF.set_index('name') for iDF in (df1, df2, df3)),
    axis=1, 
    join='inner'
).reset_index()

其中df1, df2和df3定义为John Galt的答案:

import pandas as pd
df1 = pd.DataFrame(np.array([
    ['a', 5, 9],
    ['b', 4, 61],
    ['c', 24, 9]]),
    columns=['name', 'attr11', 'attr12']
)
df2 = pd.DataFrame(np.array([
    ['a', 5, 19],
    ['b', 14, 16],
    ['c', 4, 9]]),
    columns=['name', 'attr21', 'attr22']
)
df3 = pd.DataFrame(np.array([
    ['a', 15, 49],
    ['b', 4, 36],
    ['c', 14, 9]]),
    columns=['name', 'attr31', 'attr32']
)

不需要多索引来执行连接操作。 我们只需要正确地设置索引列来执行连接操作(例如哪个命令df.set_index('Name'))

默认情况下，连接操作在索引上执行。 在本例中，只需指定Name列对应于索引。 下面是一个例子

教程可能会有用。

# Simple example where dataframes index are the name on which to perform
# the join operations
import pandas as pd
import numpy as np
name = ['Sophia' ,'Emma' ,'Isabella' ,'Olivia' ,'Ava' ,'Emily' ,'Abigail' ,'Mia']
df1 = pd.DataFrame(np.random.randn(8, 3), columns=['A','B','C'], index=name)
df2 = pd.DataFrame(np.random.randn(8, 1), columns=['D'],         index=name)
df3 = pd.DataFrame(np.random.randn(8, 2), columns=['E','F'],     index=name)
df = df1.join(df2)
df = df.join(df3)

# If you have a 'Name' column that is not the index of your dataframe,
# one can set this column to be the index
# 1) Create a column 'Name' based on the previous index
df1['Name'] = df1.index
# 1) Select the index from column 'Name'
df1 = df1.set_index('Name')

# If indexes are different, one may have to play with parameter how
gf1 = pd.DataFrame(np.random.randn(8, 3), columns=['A','B','C'], index=range(8))
gf2 = pd.DataFrame(np.random.randn(8, 1), columns=['D'], index=range(2,10))
gf3 = pd.DataFrame(np.random.randn(8, 2), columns=['E','F'], index=range(4,12))

gf = gf1.join(gf2, how='outer')
gf = gf.join(gf3, how='outer')

df1 = pd.DataFrame(np.array([
    ['a', 5, 9],
    ['b', 4, 61],
    ['c', 24, 9]]),
    columns=['name', 'attr11', 'attr12']
)
df2 = pd.DataFrame(np.array([
    ['a', 5, 19],
    ['d', 14, 16]]

),
    columns=['name', 'attr21', 'attr22']
)
df3 = pd.DataFrame(np.array([
    ['a', 15, 49],
    ['c', 4, 36],
    ['d', 14, 9]]),
    columns=['name', 'attr31', 'attr32']
)
df4 = pd.DataFrame(np.array([
    ['a', 15, 49],
    ['c', 4, 36],
    ['c', 14, 9]]),
    columns=['name', 'attr41', 'attr42']
)

加入列表数据框架的三种方法

pandas.concat

dfs = [df1, df2, df3]
dfs = [df.set_index('name') for df in dfs]
# cant not run if index not unique 
dfs = pd.concat(dfs, join='outer', axis = 1) 

functools.reduce

dfs = [df1, df2, df3, df4]
# still run with index not unique 
import functools as ft
df_final = ft.reduce(lambda left, right: pd.merge(left, right, on='name', how = 'outer'), dfs)

加入

# cant not run if index not unique 
dfs = [df1, df2, df3]
dfs = [df.set_index('name') for df in dfs]
dfs[0].join(dfs[1:], how = 'outer')

下面是一个合并数据帧字典的方法，同时保持列名与字典同步。如果需要，它还会填充缺失的值:

这是合并数据帧字典的函数

def MergeDfDict(dfDict, onCols, how='outer', naFill=None):
  keys = dfDict.keys()
  for i in range(len(keys)):
    key = keys[i]
    df0 = dfDict[key]
    cols = list(df0.columns)
    valueCols = list(filter(lambda x: x not in (onCols), cols))
    df0 = df0[onCols + valueCols]
    df0.columns = onCols + [(s + '_' + key) for s in valueCols] 

    if (i == 0):
      outDf = df0
    else:
      outDf = pd.merge(outDf, df0, how=how, on=onCols)   

  if (naFill != None):
    outDf = outDf.fillna(naFill)

  return(outDf)

好的，让我们生成数据并测试:

def GenDf(size):
  df = pd.DataFrame({'categ1':np.random.choice(a=['a', 'b', 'c', 'd', 'e'], size=size, replace=True),
                      'categ2':np.random.choice(a=['A', 'B'], size=size, replace=True), 
                      'col1':np.random.uniform(low=0.0, high=100.0, size=size), 
                      'col2':np.random.uniform(low=0.0, high=100.0, size=size)
                      })
  df = df.sort_values(['categ2', 'categ1', 'col1', 'col2'])
  return(df)


size = 5
dfDict = {'US':GenDf(size), 'IN':GenDf(size), 'GER':GenDf(size)}   
MergeDfDict(dfDict=dfDict, onCols=['categ1', 'categ2'], how='outer', naFill=0)

可以使用.join()函数将三者连接在一起。

假设你有三个数据框架 Df1 df2 df3。 要将这些数据加入到一个数据框架中，你可以:

df = df1.join(df2).join(df3)

这是我发现的完成这项任务最简单的方法。