You need mask:

sample['PR'] = sample['PR'].mask(sample['PR'] < 90, np.nan)

Another solution with loc and boolean indexing:

sample.loc[sample['PR'] < 90, 'PR'] = np.nan

Sample:

import pandas as pd
import numpy as np

sample = pd.DataFrame({'PR':[10,100,40] })
print (sample)
    PR
0   10
1  100
2   40

sample['PR'] = sample['PR'].mask(sample['PR'] < 90, np.nan)
print (sample)
      PR
0    NaN
1  100.0
2    NaN

sample.loc[sample['PR'] < 90, 'PR'] = np.nan
print (sample)
      PR
0    NaN
1  100.0
2    NaN

EDIT:

Solution with apply:

sample['PR'] = sample['PR'].apply(lambda x: np.nan if x < 90 else x)

Timings len(df)=300k:

sample = pd.concat([sample]*100000).reset_index(drop=True)

In [853]: %timeit sample['PR'].apply(lambda x: np.nan if x < 90 else x)
10 loops, best of 3: 102 ms per loop

In [854]: %timeit sample['PR'].mask(sample['PR'] < 90, np.nan)
The slowest run took 4.28 times longer than the fastest. This could mean that an intermediate result is being cached.
100 loops, best of 3: 3.71 ms per loop

Given a sample dataframe df as:

   a  b
0  1  2
1  2  3
2  3  4
3  4  5

what you want is:

df['a'] = df['a'].apply(lambda x: x + 1)

that returns:

   a  b
0  2  2
1  3  3
2  4  4
3  5  5

You can use function map and select by function np.where more info

print df
#     a     b
#0  aaa  rrrr
#1   bb     k
#2  ccc     e
#condition if condition is True then len column a else column b
df['c'] = np.where(df['a'].map(len) > df['b'].map(len), df['a'].map(len), df['b'].map(len))
print df
#     a     b  c
#0  aaa  rrrr  4
#1   bb     k  2
#2  ccc     e  3

Next solution is with function apply with parameter axis=1:

axis = 1 or ‘columns’: apply function to each row

df['c'] = df.apply(lambda x: max(len(x['a']), len(x['b'])), axis=1)