Python-熊猫，将长列拆分为多列

Question

Given the following DataFrame:

>>> pd.DataFrame(data=[['a',1],['a',2],['b',3],['b',4],['c',5],['c',6],['d',7],['d',8],['d',9],['e',10]],columns=['key','value'])
  key  value
0   a      1
1   a      2
2   b      3
3   b      4
4   c      5
5   c      6
6   d      7
7   d      8
8   d      9
9   e     10

I'm looking for a method that will change the structure based on the key value, like so:

   a  b  c  d   e
0  1  3  5  7  10
1  2  4  6  8  10 <- 10 is duplicated
2  2  4  6  9  10 <- 10 is duplicated

The result row number is as the longest group count (d in the above example) and the missing values are duplicates of the last available value.

Answer 1

Create MultiIndex by set_index with counter column by cumcount , reshape by unstack , repalce missing values by last non missing ones with ffill and last converting all data to integer s if necessary:

df = df.set_index([df.groupby('key').cumcount(),'key'])['value'].unstack().ffill().astype(int)

Another solution with custom lambda function:

df = (df.groupby('key')['value']
        .apply(lambda x: pd.Series(x.values))
        .unstack(0)
        .ffill()
        .astype(int))

---

print (df)
key  a  b  c  d   e
0    1  3  5  7  10
1    2  4  6  8  10
2    2  4  6  9  10

Answer 2

Using pivot , with groupby + cumcount

df.assign(key2=df.groupby('key').cumcount()).pivot('key2','key','value').ffill().astype(int)
Out[214]: 
key   a  b  c  d   e
key2                
0     1  3  5  7  10
1     2  4  6  8  10
2     2  4  6  9  10