Proper way to use iloc in Pandas

Question

I have the following dataframe df:

print(df)

    Food         Taste
0   Apple        NaN
1   Banana       NaN
2   Candy        NaN
3   Milk         NaN
4   Bread        NaN
5   Strawberry   NaN

I am trying to replace values in a range of rows using iloc:

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

But it returned the following SettingWithCopyWarning message:

SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame

So, I found this Stackoverflow page and tried this:

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

Unfortunately, it returned the following error:

ValueError: Can only index by location with a [integer, integer slice (START point is INCLUDED, END point is EXCLUDED), listlike of integers, boolean array]

What would be the proper way to use iloc in this situation? Also, is there a way to combine these two lines above?

Answer 1

Purely integer-location based indexing for selection by position.. eg :-

lang_sets = {}
lang_sets['en'] = train[train.lang == 'en'].iloc[:,:-1]
lang_sets['ja'] = train[train.lang == 'ja'].iloc[:,:-1]
lang_sets['de'] = train[train.lang == 'de'].iloc[:,:-1]

Answer 2

.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.

Answer 3

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

Answer 4

I prefer to use .loc in such cases, and explicitly use the index of the DataFrame if you want to select on position:

df.loc[df.index[0:2], 'Taste'] = 'good'
df.loc[df.index[2:6], 'Taste'] = 'bad'