将元组列表转换为dataframe - 其中元组的第一个元素是列名

Question

I have a list of tuples in the format:

tuples = [('a',1,10,15),('b',11,0,3),('c',7,19,2)]  # etc.

I wish to store the data in a DataFrame with the format:

      a       b     c      ...  

0     1       11     7     ...   
1     10      0      19    ...  
2     15      3      2     ...   

Where the first element of the tuple is what I wish to be the column name.

I understand that if I can achieve what I want by running:

df = pd.DataFrame(tuples)
df = df.T
df.columns = df.iloc[0]
df = df[1:]

But it seems to me like it should be more straightforward than this. Is this a more pythonic way of solving this?

Answer 1

Here's one way

In [151]: pd.DataFrame({x[0]:x[1:] for x in tuples})
Out[151]:
    a   b   c
0   1  11   7
1  10   0  19
2  15   3   2

Answer 2

You can use dictionary comprehension, like:

pd.DataFrame({k:v for k,*v in tuples})

in python-3.x , or:

pd.DataFrame({t[0]: t[1:] for t in tuples})

in python-2.7 .

which generates:

>>> pd.DataFrame({k:v for k,*v in tuples})
    a   b   c
0   1  11   7
1  10   0  19
2  15   3   2

The columns will be sorted alphabetically .

If you want the columns to be sorted like the original content , you can use the columns parameter:

pd.DataFrame({k:v for k,*v in tuples})

again in python-3.x , or for python-2.7 :

pd.DataFrame({t[0]: t[1:] for t in tuples})

We can shorten this a bit into:

from operator import itemgetter

pd.DataFrame({t[0]: t[1:] for t in tuples},columns=tuples)