HDF5.remove with where removing all rows
#### when I use 'in' condition in HDFStore.remove, it removes all data in the table instead of selected rows

```python
In[0]:
store = pd.HDFStore("d:/org.h5")
a = store.get(table["table_name"])
a.index
​
Out[0]:
Index(['{7D23701A-991B-11E7-977E-448A5B7647CF}',
       '{7D23701E-991B-11E7-977E-448A5B7647CF}',
       '{7D237020-991B-11E7-977E-448A5B7647CF}',
       ...
       '{9E1257B7-99E4-11E7-B36E-448A5B7647CF}',
       '{9E1257B9-99E4-11E7-B36E-448A5B7647CF}',
       '{9E1257CB-99E4-11E7-B36E-448A5B7647CF}'],
      dtype='object', name='object_id', length=6710)

In [1]:
a[a['trade_dt'] == '2017-09-14'].index

Out[1]:
Index(['{7D23701A-991B-11E7-977E-448A5B7647CF}',
       '{7D23701E-991B-11E7-977E-448A5B7647CF}',
       '{7D237020-991B-11E7-977E-448A5B7647CF}',
       ...
       '{7D236E8C-991B-11E7-977E-448A5B7647CF}',
       '{7D236E9A-991B-11E7-977E-448A5B7647CF}',
       '{7D236EA0-991B-11E7-977E-448A5B7647CF}'],
      dtype='object', name='object_id', length=3353)

In [2]:
store.select(table["table_name"], where="index in %s" % list(a[a['trade_dt'] == '2017-09-14'].index))

Out[2]:
(3353, 22)

In [3]:
store.remove(table["table_name"], where="index in %s" % list(a[a['trade_dt'] == '2017-09-14'].index))

Out[3]:
6710

```
#### Problem description

The “where" condition in HDFStore.select returns selected 3353 rows. I supposed using same condition with "where" in HDFStore.remove should do the same which is remove this 3353 rows. But it remove 6710 rows which is all lines in this table.

I tested with only one single row in the "index in [list]" where condition. It removes one row which is I expected.

```python

In [4]:
[a[a['trade_dt'] == '2017-09-14'].index[0]]

Out[4]:
['{7D23701A-991B-11E7-977E-448A5B7647CF}']

In [5]:
store.select(table["table_name"], where="index in %s" % [a[a['trade_dt'] == '2017-09-14'].index[0]]).shape

Out[5]:
(1, 22)

In [6]:
store.remove(table["table_name"], where="index in %s" % [a[a['trade_dt'] == '2017-09-14'].index[0]])

Out[6]:
1

```

I tested with equivalent condition ops such as "=" returns the same

#### Output of ``pd.show_versions()``

<details>

'0.20.3'

</details>
