爆炸衣衫襤褸的列表中的一列數據框入鑑於由可變大小的列表的列的數據幀多列

Question

：

    Col1 
0   [SF, NYG, 123] 
1 [SF, NYG, test, test] 
2   [SF, NYG, foo] 
3    [SF, NYG] 
4   [SF, NYG, 45] 
5    [SF, NYG] 
6   [SF, NYG, 32] 

我怎樣才能將其轉換爲多列的數據幀？我想空值有NaN，像這樣：

Col1 Col2 Col3 Col4 
0 SF NYG 123 NaN 
1 SF NYG test test 
2 SF NYG foo NaN 
3 SF NYG NaN NaN 
4 SF NYG 45 NaN 
5 SF NYG NaN NaN 
6 SF NYG 32 NaN 

---

我能拿出

df_new = pd.DataFrame(df.Col1.tolist()).applymap(lambda x: np.nan if not x else x) 

但我無法找到一個方法來重命名優雅的列。

Answer 1

您可以使用DataFrame構造與values爲numpy array從list。然後更換None到NaN和重命名列，最後add_prefix

df_new = pd.DataFrame(df.Col1.values.tolist()) 
      .fillna(np.nan) 
      .rename(columns = lambda x: x + 1) 
      .add_prefix('Col') 
print (df_new) 
    Col1 Col2 Col3 Col4 
0 SF NYG 123 NaN 
1 SF NYG test test 
2 SF NYG foo NaN 
3 SF NYG NaN NaN 
4 SF NYG 45 NaN 
5 SF NYG NaN NaN 
6 SF NYG 32 NaN 

時序：

#700 
df = pd.concat([df]*100).reset_index(drop=True) 

#Jez 
In [10]: %timeit (pd.DataFrame(df.Col1.values.tolist())) 
1000 loops, best of 3: 694 µs per loop 

#cᴏʟᴅsᴘᴇᴇᴅ 
In [11]: %timeit (pd.DataFrame(df.Col1.tolist())) 
1000 loops, best of 3: 705 µs per loop 

#Wen 
In [12]: %timeit (df.Col1.apply(lambda x: ','.join(str(y) for y in x)).str.split(',', expand=True)) 
100 loops, best of 3: 3.51 ms per loop 

#slowier 
In [13]: %timeit (df.Col1.apply(pd.Series)) 
10 loops, best of 3: 159 ms per loop 

---

#7k 
df = pd.concat([df]*1000).reset_index(drop=True) 

#jez 
In [30]: %timeit (pd.DataFrame(df.Col1.values.tolist())) 
1000 loops, best of 3: 1.26 ms per loop 

#cᴏʟᴅsᴘᴇᴇᴅ 
In [31]: %timeit (pd.DataFrame(df.Col1.tolist())) 
1000 loops, best of 3: 1.37 ms per loop 

#Wen 
In [32]: %timeit (df.Col1.apply(lambda x: ','.join(str(y) for y in x)).str.split(',', expand=True)) 
10 loops, best of 3: 29 ms per loop 

#very slow, the best use only in small dataframes 
In [33]: %timeit (df.Col1.apply(pd.Series)) 
1 loop, best of 3: 1.58 s per loop 

---

#700k 
df = pd.concat([df]*100000).reset_index(drop=True) 

#jez 
In [40]: %timeit (pd.DataFrame(df.Col1.values.tolist())) 
10 loops, best of 3: 80.3 ms per loop 

#cᴏʟᴅsᴘᴇᴇᴅ 
In [41]: %timeit (pd.DataFrame(df.Col1.tolist())) 
10 loops, best of 3: 90.5 ms per loop 

#Wen 
In [42]: %timeit (df.Col1.apply(lambda x: ','.join(str(y) for y in x)).str.split(',', expand=True)) 
1 loop, best of 3: 2.91 s per loop 

#extremely slow 
In [3]: %timeit (df.Col1.apply(pd.Series)) 
1 loop, best of 3: 3min 58s per loop 

Answer 2

或者你也可以做到這一點

df.var1.apply(lambda x: ','.join(str(y) for y in x)).str.split(',', expand=True).fillna(value=np.nan)