python-用单个行连接两个数据框熊猫

我有一个数据框df看起来像：

   one  three  two
0  1.0   10.0  4.0
1  2.0    3.0  3.0
2  3.0   22.0  2.0
3  4.0    1.0  1.0

我还有另一个单行数据框df2,如下所示：

     a    b    m    u
0  1.0  2.0  1.0  4.0

我想将两者连接起来,最后得到：

   one  three  two    a    b    m    u
0  1.0   10.0  4.0  1.0  2.0  1.0  4.0
1  2.0    3.0  3.0  1.0  2.0  1.0  4.0
2  3.0   22.0  2.0  1.0  2.0  1.0  4.0
3  4.0    1.0  1.0  1.0  2.0  1.0  4.0

我试过了：

df3 = pd.concat([df, df2], axis=1, ignore_index=True)

     0     1    2    3    4    5    6
0  1.0  10.0  4.0  1.0  2.0  1.0  4.0
1  2.0   3.0  3.0  NaN  NaN  NaN  NaN
2  3.0  22.0  2.0  NaN  NaN  NaN  NaN
3  4.0   1.0  1.0  NaN  NaN  NaN  NaN

错误答案…

我该如何解决呢？

非常感谢.

解决方法:

我认为您可以将numpy.tile用于重复数据：

df2 = pd.DataFrame(np.tile(df2.values, len(df.index)).reshape(-1,len(df2.columns)), 
                   columns=df2.columns)
print (df2)
     a    b    m    u
0  1.0  2.0  1.0  4.0
1  1.0  2.0  1.0  4.0
2  1.0  2.0  1.0  4.0
3  1.0  2.0  1.0  4.0

df3 = df.join(df2)
print (df3)
   one  three  two    a    b    m    u
0  1.0   10.0  4.0  1.0  2.0  1.0  4.0
1  2.0    3.0  3.0  1.0  2.0  1.0  4.0
2  3.0   22.0  2.0  1.0  2.0  1.0  4.0
3  4.0    1.0  1.0  1.0  2.0  1.0  4.0

或改进了John Galt solution-仅替换了df2列的NaN：

df3 = df.join(df2)
df3[df2.columns] = df3[df2.columns].ffill()
print (df3)
   one  three  two    a    b    m    u
0  1.0   10.0  4.0  1.0  2.0  1.0  4.0
1  2.0    3.0  3.0  1.0  2.0  1.0  4.0
2  3.0   22.0  2.0  1.0  2.0  1.0  4.0
3  4.0    1.0  1.0  1.0  2.0  1.0  4.0

由iloc创建的assign by Series的另一种解决方案,但列名称必须为字符串：

df3 = df.assign(**df2.iloc[0])
print (df3)
   one  three  two    a    b    m    u
0  1.0   10.0  4.0  1.0  2.0  1.0  4.0
1  2.0    3.0  3.0  1.0  2.0  1.0  4.0
2  3.0   22.0  2.0  1.0  2.0  1.0  4.0
3  4.0    1.0  1.0  1.0  2.0  1.0  4.0

时间：

np.random.seed(44)
N = 1000000

df = pd.DataFrame(np.random.random((N,5)), columns=list('ABCDE'))

df2 = pd.DataFrame(np.random.random((1, 50)))
df2.columns = 'a' + df2.columns.astype(str)


In [369]: %timeit df.join(pd.DataFrame(np.tile(df2.values, len(df.index)).reshape(-1,len(df2.columns)), columns=df2.columns))
1 loop, best of 3: 897 ms per loop

In [370]: %timeit df.assign(**df2.iloc[0])
1 loop, best of 3: 467 ms per loop

In [371]: %timeit df.assign(key=1).merge(df2.assign(key=1), on='key').drop('key',axis=1)
1 loop, best of 3: 1.55 s per loop

In [372]: %%timeit
     ...: df3 = df.join(df2)
     ...: df3[df2.columns] = df3[df2.columns].ffill()
     ...: 
1 loop, best of 3: 1.9 s per loop

标签：pandas,concatenation,python
来源： https://codeday.me/bug/20191025/1931158.html