Pandas (and numpy) allow for boolean indexing, which will be much more efficient:

In [11]: df.loc[df['col1'] >= 1, 'col1']
Out[11]: 
1    1
2    2
Name: col1

In [12]: df[df['col1'] >= 1]
Out[12]: 
   col1  col2
1     1    11
2     2    12

In [13]: df[(df['col1'] >= 1) & (df['col1'] <=1 )]
Out[13]: 
   col1  col2
1     1    11

If you want to write helper functions for this, consider something along these lines:

In [14]: def b(x, col, op, n): 
             return op(x[col],n)

In [15]: def f(x, *b):
             return x[(np.logical_and(*b))]

In [16]: b1 = b(df, 'col1', ge, 1)

In [17]: b2 = b(df, 'col1', le, 1)

In [18]: f(df, b1, b2)
Out[18]: 
   col1  col2
1     1    11

Update: pandas 0.13 has a query method for these kind of use cases, assuming column names are valid identifiers the following works (and can be more efficient for large frames as it uses numexpr behind the scenes):

In [21]: df.query('col1 <= 1 & 1 <= col1')
Out[21]:
   col1  col2
1     1    11

Use () because operator precedence:

temp2 = df[~df["Def"] & (df["days since"] > 7) & (df["bin"] == 3)]

Alternatively, create conditions on separate rows:

cond1 = df["bin"] == 3    
cond2 = df["days since"] > 7
cond3 = ~df["Def"]

temp2 = df[cond1 & cond2 & cond3]

Sample:

df = pd.DataFrame({'Def':[True] *2 + [False]*4,
                   'days since':[7,8,9,14,2,13],
                   'bin':[1,3,5,3,3,3]})

print (df)
     Def  bin  days since
0   True    1           7
1   True    3           8
2  False    5           9
3  False    3          14
4  False    3           2
5  False    3          13


temp2 = df[~df["Def"] & (df["days since"] > 7) & (df["bin"] == 3)]
print (temp2)
     Def  days since  bin
3  False          14    3
5  False          13    3

Another idea is use functions like Series.lt Series.gt Series.le Series.ge Series.ne:

temp2 = df[~df["Def"] & df["days since"].gt(7) & df["bin"].eq(3)]
print (temp2)
     Def  days since  bin
3  False          14    3
5  False          13    3