我有一个这样的数据框“ bb”:
Response Unique Count
I love it so much! 246_0 1
This is not bad, but can be better. 246_1 2
Well done, let's do it. 247_0 1
如果count大于1,我想分割字符串并使数据框“ bb”变成这样:(我期望的结果)
Response Unique
I love it so much! 246_0
This is not bad 246_1_0
but can be better. 246_1_1
Well done, let's do it. 247_0
我的代码:
bb = DataFrame(bb[bb['Count'] > 1].Response.str.split(',').tolist(), index=bb[bb['Count'] > 1].Unique).stack()
bb = bb.reset_index()[[0, 'Unique']]
bb.columns = ['Response','Unique']
bb=bb.replace('', np.nan)
bb=bb.dropna()
print(bb)
但是结果是这样的:
Response Unique
0 This is not bad 246_1
1 but can be better. 246_1
在这种情况下如何保留原始数据框?
答案 0 :(得分:3)
首先将每个条件的值仅与新辅助程序Series
分开,然后仅对每个重复的索引值按GroupBy.cumcount
按Index.duplicated
来添加计数器值:
s = df.loc[df.pop('Count') > 1, 'Response'].str.split(',', expand=True).stack()
df1 = df.join(s.reset_index(drop=True, level=1).rename('Response1'))
df1['Response'] = df1.pop('Response1').fillna(df1['Response'])
mask = df1.index.duplicated(keep=False)
df1.loc[mask, 'Unique'] += df1[mask].groupby(level=0).cumcount().astype(str).radd('_')
df1 = df1.reset_index(drop=True)
print (df1)
Response Unique
0 I love it so much! 246_0
1 This is not bad 246_1_0
2 but can be better. 246_1_1
3 Well done! 247_0
编辑:如果需要_0
的所有其他值,请删除掩码:
s = df.loc[df.pop('Count') > 1, 'Response'].str.split(',', expand=True).stack()
df1 = df.join(s.reset_index(drop=True, level=1).rename('Response1'))
df1['Response'] = df1.pop('Response1').fillna(df1['Response'])
df1['Unique'] += df1.groupby(level=0).cumcount().astype(str).radd('_')
df1 = df1.reset_index(drop=True)
print (df1)
Response Unique
0 I love it so much! 246_0_0
1 This is not bad 246_1_0
2 but can be better. 246_1_1
3 Well done! 247_0_0
答案 1 :(得分:1)
我们可以逐步解决以下问题:
groupby
,并使用cumcount
获取正确的unique
列值。 concat
将数据帧重新组合在一起。df1 = df[df['Count'].ge(2)] # all rows which have a count 2 or higher
df2 = df[df['Count'].eq(1)] # all rows which have count 1
df1 = explode_str(df1, 'Response', ',') # explode the string to rows on comma delimiter
# Create the correct unique column
df1['Unique'] = df1['Unique'] + '_' + df1.groupby(df1.index).cumcount().astype(str)
df = pd.concat([df1, df2]).sort_index().drop('Count', axis=1).reset_index(drop=True)
Response Unique
0 I love it so much! 246_0
1 This is not bad 246_1_0
2 but can be better. 246_1_1
3 Well done! 247_0
链接答案中使用的功能:
def explode_str(df, col, sep):
s = df[col]
i = np.arange(len(s)).repeat(s.str.count(sep) + 1)
return df.iloc[i].assign(**{col: sep.join(s).split(sep)})