Proper way to use iloc in Pandas

You can use Index.get_loc for position of column Taste, because DataFrame.iloc select by positions:

#return second position (python counts from 0, so 1)
print (df.columns.get_loc('Taste'))
1

df.iloc[0:2, df.columns.get_loc('Taste')] = 'good'
df.iloc[2:6, df.columns.get_loc('Taste')] = 'bad'
print (df)
         Food Taste
0       Apple  good
1      Banana  good
2       Candy   bad
3        Milk   bad
4       Bread   bad
5  Strawberry   bad

Possible solution with ix is not recommended because deprecate ix in next version of pandas:

df.ix[0:2, 'Taste'] = 'good'
df.ix[2:6, 'Taste'] = 'bad'
print (df)
         Food Taste
0       Apple  good
1      Banana  good
2       Candy   bad
3        Milk   bad
4       Bread   bad
5  Strawberry   bad

.iloc uses integer location, whereas .loc uses name. Both options also take both row AND column identifiers (for DataFrames). Your inital code didn't work because you didn't specify within the .iloc call which column you're selecting. The second code line you tried didn't work because you mixed integer location with column name, and .iloc only accepts integer location. If you don't know the column integer location, you can use Index.get_loc in place as suggested above. Otherwise, use the integer position, in this case 1.

df.iloc[0:2, df.columns.get_loc('Taste')] = 'good'
df.iloc[2:6, df.columns.get_loc('Taste')] = 'bad'

is equal to:

df.iloc[0:2, 1] = 'good'
df.iloc[2:6, 1] = 'bad'

in this particular situation.