Just compare the column with that value:

In [9]: df = pandas.DataFrame([1,2,3,4], columns=["data"])

In [10]: df
Out[10]: 
   data
0     1
1     2
2     3
3     4

In [11]: df["desired"] = df["data"] > 2.5
In [11]: df
Out[12]: 
   data desired
0     1   False
1     2   False
2     3    True
3     4    True

Depending on the complexity there is probably a more efficient answer here. I use np.where to get your answer.

 import numpy as np
 df['conditional'] = np.where(
      (df['conditional'] == 'A'), # Condition
      'a', # Value if true
      df['conditional'] # Value if false
      # /\ this is equal to the original value so it has no effect
 )

Documentation: https://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html

I do not recommend the use of apply here: it should be avoided if there are better alternatives.

For example, if you are performing the following operation on a Series:

if cond1:
    exp1
elif cond2:
    exp2
else:
    exp3

This is usually a good use case for np.where or np.select.

numpy.where

The if else chain above can be written using

np.where(cond1, exp1, np.where(cond2, exp2, ...))

np.where allows nesting. With one level of nesting, your problem can be solved with,

df['three'] = (
    np.where(
        df['one'] < 2, 
        df['one'] * 10, 
        np.where(df['one'] < 4, df['one'] ** 2, df['one'] + 10))
df

   one  two  three
0    1    6     10
1    2    7      4
2    3    8      9
3    4    9     14
4    5   10     15

numpy.select

Allows for flexible syntax and is easily extensible. It follows the form,

np.select([cond1, cond2, ...], [exp1, exp2, ...])

Or, in this case,

np.select([cond1, cond2], [exp1, exp2], default=exp3)

df['three'] = (
    np.select(
        condlist=[df['one'] < 2, df['one'] < 4], 
        choicelist=[df['one'] * 10, df['one'] ** 2], 
        default=df['one'] + 10))
df

   one  two  three
0    1    6     10
1    2    7      4
2    3    8      9
3    4    9     14
4    5   10     15

and/or (similar to the if/else)

Similar to if-else, requires the lambda:

df['three'] = df["one"].apply(
    lambda x: (x < 2 and x * 10) or (x < 4 and x ** 2) or x + 10) 

df
   one  two  three
0    1    6     10
1    2    7      4
2    3    8      9
3    4    9     14
4    5   10     15

List Comprehension

Loopy solution that is still faster than apply.

df['three'] = [x*10 if x<2 else (x**2 if x<4 else x+10) for x in df['one']]
# df['three'] = [
#    (x < 2 and x * 10) or (x < 4 and x ** 2) or x + 10) for x in df['one']
# ]
df
   one  two  three
0    1    6     10
1    2    7      4
2    3    8      9
3    4    9     14
4    5   10     15

All you want is to trim the strings to atmost 5 places. You can do df.ZipCode.str[:5]?

In [78]: df
Out[78]:
     ZipCode
0  123456789
1        123
2          0

For, 5 places zipcodes

In [79]: df.ZipCode.str[:5]
Out[79]:
0    12345
1      123
2        0
Name: ZipCode, dtype: object

For, the extra (>5) characters part in zipcodes

In [80]: df.ZipCode.str[5:]
Out[80]:
0    6789
1
2
Name: ZipCode, dtype: object

Solution

use a vectorized approach

frame['d'] = frame.b + (frame.c > 0) * (frame.c - frame.b)

Explanation

This is derived from the sum of

(frame.c > 0) * frame.c  # frame.c if positive

Plus

(frame.c <= 0) * frame.b  # frame.b if c is not positive

However

(frame.c <=0 )

is equivalent to

(1 - frame.c > 0)

and when combined you get

frame['d'] = frame.b + (frame.c > 0) * (frame.c - frame.b)