You can replace this just for that column using replace:

df['workclass'].replace('?', np.NaN)

or for the whole df:

df.replace('?', np.NaN)

UPDATE

OK I figured out your problem, by default if you don't pass a separator character then read_csv will use commas ',' as the separator.

Your data and in particular one example where you have a problematic line:

54, ?, 180211, Some-college, 10, Married-civ-spouse, ?, Husband, Asian-Pac-Islander, Male, 0, 0, 60, South, >50K

has in fact a comma and a space as the separator so when you passed the na_value=['?'] this didn't match because all your values have a space character in front of them all which you can't observe.

if you change your line to this:

rawfile = pd.read_csv(filename, header=None, names=DataLabels, sep=',\s', na_values=["?"])

then you should find that it all works:

27      54               NaN  180211  Some-college             10 

DataFrame.fillna() or Series.fillna() will do this for you.

Example:

In [7]: df
Out[7]: 
          0         1
0       NaN       NaN
1 -0.494375  0.570994
2       NaN       NaN
3  1.876360 -0.229738
4       NaN       NaN

In [8]: df.fillna(0)
Out[8]: 
          0         1
0  0.000000  0.000000
1 -0.494375  0.570994
2  0.000000  0.000000
3  1.876360 -0.229738
4  0.000000  0.000000

To fill the NaNs in only one column, select just that column.

In [12]: df[1] = df[1].fillna(0)

In [13]: df
Out[13]: 
          0         1
0       NaN  0.000000
1 -0.494375  0.570994
2       NaN  0.000000
3  1.876360 -0.229738
4       NaN  0.000000

Or you can use the built in column-specific functionality:

df = df.fillna({1: 0})

Actually in later versions of pandas this will give a TypeError:

df.replace('-', None)
TypeError: If "to_replace" and "value" are both None then regex must be a mapping

You can do it by passing either a list or a dictionary:

In [11]: df.replace('-', df.replace(['-'], [None]) # or .replace('-', {0: None})
Out[11]:
      0
0  None
1     3
2     2
3     5
4     1
5    -5
6    -1
7  None
8     9

But I recommend using NaNs rather than None:

In [12]: df.replace('-', np.nan)
Out[12]:
     0
0  NaN
1    3
2    2
3    5
4    1
5   -5
6   -1
7  NaN
8    9

dataframe

You can use pd.DataFrame.mask:

df.mask((df >= -200) & (df <= -100), inplace=True)

This method replaces elements identified by True values in a Boolean array with a specified value, defaulting to NaN if a value is not specified.

Equivalently, use pd.DataFrame.where with the reverse condition:

df.where((df < -200) | (df > -100), inplace=True)

series

As with many methods, Pandas helpfully includes versions which work with series rather than an entire dataframe. So, for a column df['A'], you can use pd.Series.mask with pd.Series.between:

df['A'].mask(df['A'].between(-200, -100), inplace=True)

For chaining, note inplace=False by default, so you can also use:

df['A'] = df['A'].mask(df['A'].between(-200, -100))

Randomly replace values in a numpy array

# The dataset
data = pd.read_csv('iris.data')
mat = data.iloc[:,:4].as_matrix()

Set the number of values to replace. For example 20%:

# Edit: changed len(mat) for mat.size
prop = int(mat.size * 0.2)

Randomly choose indices of the numpy array:

i = [random.choice(range(mat.shape[0])) for _ in range(prop)]
j = [random.choice(range(mat.shape[1])) for _ in range(prop)]

Change values with NaN

mat[i,j] = np.NaN

Dropout for any array dimension

Another way to do that with an array of more than 2 dimensions would be to use the numpy.put() function:

import numpy as np
import random
from sklearn import datasets

data = datasets.load_iris()['data']

def dropout(a, percent):
    # create a copy
    mat = a.copy()
    # number of values to replace
    prop = int(mat.size * percent)
    # indices to mask
    mask = random.sample(range(mat.size), prop)
    # replace with NaN
    np.put(mat, mask, [np.NaN]*len(mask))
    return mat

This function returns a modified array:

modified = dropout(data, 0.2)

We can verify that the correct number of values have been modified:

np.sum(np.isnan(modified))/float(data.size)

[out]:

0.2

There is no one size fits all. So you cannot assume that one technique will work the best for all the datasets.

That being said the goal of imputing missing values is to ensure that after imputation, the distribution of the column does not change. So if you have a feature that follows a left skewed distribution, then after imputation the distribution should not change much.

Following this logic use multiple imputation techniques to see which one retains the original distribution of the feature you are imputing the values for.