合并以相同字母开头的pandas DataFrame列 [英] Merge pandas DataFrame columns starting with the same letters

问题描述

假设我有一个DataFrame:

>>> df = pd.DataFrame({'a1':[1,2],'a2':[3,4],'b1':[5,6],'b2':[7,8],'c':[9,0]})
>>> df
   a1  a2  b1  b2  c
0   1   3   5   7  9
1   2   4   6   8  0
>>> 

我想合并(也许不是合并，而是将其名称的第一个字母相等的列合并，例如a1和a2以及其他列)，但是正如我们所看到的，有一个列，它本身没有任何其他类似的列，因此，我希望它们不要抛出错误，而要向它们添加NaN.

And I want to merge (maybe not merge, but concatenate) the columns where their name's first letter are equal, such as a1 and a2 and others... but as we see, there is a c column which is by itself without any other similar ones, therefore I want them to not throw errors, instead add NaNs to them.

我想以某种方式合并，它将宽范围的DataFrame更改为长范围的DataFrame，基本上就像宽范围到长范围的修改一样.

I want to merge in a way that it will change a wide DataFrame into a long DataFrame, basically like a wide to long modification.

我已经有解决问题的方法，但是唯一的问题是它的效率很低，我想要一个更高效，更快速的解决方案(与我的:P不同)，我目前有一个for循环和一个try except(嗯，听起来已经很糟糕了)，例如:

I already have a solution to the problem, but only thing is that it's very inefficient, I would like a more efficient and faster solution (unlike mine :P), I currently have a for loop and a try except (ugh, sounds bad already) code such as:

>>> df2 = pd.DataFrame()
>>> for i in df.columns.str[:1].unique():
    try:
        df2[i] = df[[x for x in df.columns if x[:1] == i]].values.flatten()
    except:
        l = df[[x for x in df.columns if x[:1] == i]].values.flatten().tolist()
        df2[i] = l + [pd.np.nan] * (len(df2) - len(l))


>>> df2
   a  b    c
0  1  5  9.0
1  3  7  0.0
2  2  6  NaN
3  4  8  NaN
>>> 

我想用更好的代码获得相同的结果.

I would like to obtain the same results with better code.

推荐答案

使用字典理解:

df = pd.DataFrame({i: pd.Series(x.to_numpy().ravel()) 
                      for i, x in df.groupby(lambda x: x[0], axis=1)})
print (df)
   a  b    c
0  1  5  9.0
1  3  7  0.0
2  2  6  NaN
3  4  8  NaN

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