根据熊猫的索引范围合并列的行

时间:2019-11-25 12:48:50

标签: python pandas list list-comprehension

我有一个只有一列的数据框。

Index | column1 |
0         and
1         too
2         ask
3         the
4         but
5         hat
6         hot
7         top
8         tap

我想根据条件合并索引之间的行。例如,如果一行中有字母“ a”,则索引将为:

0, 2, 5, 8

因此,合并行:

(0, 1), (2, 3, 4), (5, 6, 7), (8)

最后输出是:

Index | column1 |
0         and, too
1         ask, the, but
2         hat, hot, top
3         tap

我尝试过的是:

[i for i in range(len(df['column1'])) if 'a' in df['column1'][i]]

给我指数:

[0, 2, 5, 8]

但是从这里卡住了。谢谢

2 个答案:

答案 0 :(得分:3)

aSeries.str.contains进行比较,并按Series.cumsum创建组,然后通过过滤a删除可能包含非g[g > 0]值的第一个组,最后用join

g = df['column1'].str.contains('a').cumsum()

df = df.groupby(g[g > 0])['column1'].apply(', '.join).reset_index(drop=True).to_frame()
print (df)
         column1
0       and, too
1  ask, the, but
2  hat, hot, top
3            tap

第一个值不包含a

print (df)
  column1
1     too
2     ask
3     the
4     but
5     hat
6     hot
7     top
8     tap

g = df['column1'].str.contains('a').cumsum()

df = df.groupby(g[g > 0])['column1'].apply(', '.join).reset_index(drop=True).to_frame()
print (df)
         column1
0  ask, the, but
1  hat, hot, top
2            tap

答案 1 :(得分:0)

stuff=["and","too","ask","the","but","hat","hot","top","tap"]

newlist=[]
collection=[]
for i in stuff:
    if "a" in i:
        if len(collection) >0:
            newlist.append(collection)
        collection=[]
    collection.append(i)
newlist.append(collection)

尝试这样的事情