Selecting pandas column by location

Two approaches that come to mind:

>>> df
          A         B         C         D
0  0.424634  1.716633  0.282734  2.086944
1 -1.325816  2.056277  2.583704 -0.776403
2  1.457809 -0.407279 -1.560583 -1.316246
3 -0.757134 -1.321025  1.325853 -2.513373
4  1.366180 -1.265185 -2.184617  0.881514
>>> df.iloc[:, 2]
0    0.282734
1    2.583704
2   -1.560583
3    1.325853
4   -2.184617
Name: C
>>> df[df.columns[2]]
0    0.282734
1    2.583704
2   -1.560583
3    1.325853
4   -2.184617
Name: C

---

Edit: The original answer suggested the use of df.ix[:,2] but this function is now deprecated. Users should switch to df.iloc[:,2].

How to select columns by position in pandas

If you want to combine column indexes you can also do this:

df_all[df_all.columns[10:21] | df_all.columns[-1:]]

Selecting multiple columns in a Pandas dataframe

The column names (which are strings) cannot be sliced in the manner you tried.

Here you have a couple of options. If you know from context which variables you want to slice out, you can just return a view of only those columns by passing a list into the __getitem__ syntax (the []'s).

df1 = df[['a', 'b']]

Alternatively, if it matters to index them numerically and not by their name (say your code should automatically do this without knowing the names of the first two columns) then you can do this instead:

df1 = df.iloc[:, 0:2] # Remember that Python does not slice inclusive of the ending index.

Additionally, you should familiarize yourself with the idea of a view into a Pandas object vs. a copy of that object. The first of the above methods will return a new copy in memory of the desired sub-object (the desired slices).

Sometimes, however, there are indexing conventions in Pandas that don't do this and instead give you a new variable that just refers to the same chunk of memory as the sub-object or slice in the original object. This will happen with the second way of indexing, so you can modify it with the .copy() method to get a regular copy. When this happens, changing what you think is the sliced object can sometimes alter the original object. Always good to be on the look out for this.

df1 = df.iloc[0, 0:2].copy() # To avoid the case where changing df1 also changes df

To use iloc, you need to know the column positions (or indices). As the column positions may change, instead of hard-coding indices, you can use iloc along with get_loc function of columns method of dataframe object to obtain column indices.

{df.columns.get_loc(c): c for idx, c in enumerate(df.columns)}

Now you can use this dictionary to access columns through names and using iloc.

How to take column-slices of dataframe in pandas

2017 Answer - pandas 0.20: .ix is deprecated. Use .loc

See the deprecation in the docs

.loc uses label based indexing to select both rows and columns. The labels being the values of the index or the columns. Slicing with .loc includes the last element.

Let's assume we have a DataFrame with the following columns:

foo, bar, quz, ant, cat, sat, dat.

# selects all rows and all columns beginning at 'foo' up to and including 'sat'
df.loc[:, 'foo':'sat']
# foo bar quz ant cat sat

.loc accepts the same slice notation that Python lists do for both row and columns. Slice notation being start:stop:step

# slice from 'foo' to 'cat' by every 2nd column
df.loc[:, 'foo':'cat':2]
# foo quz cat

# slice from the beginning to 'bar'
df.loc[:, :'bar']
# foo bar

# slice from 'quz' to the end by 3
df.loc[:, 'quz'::3]
# quz sat

# attempt from 'sat' to 'bar'
df.loc[:, 'sat':'bar']
# no columns returned

# slice from 'sat' to 'bar'
df.loc[:, 'sat':'bar':-1]
sat cat ant quz bar

# slice notation is syntatic sugar for the slice function
# slice from 'quz' to the end by 2 with slice function
df.loc[:, slice('quz',None, 2)]
# quz cat dat

# select specific columns with a list
# select columns foo, bar and dat
df.loc[:, ['foo','bar','dat']]
# foo bar dat

You can slice by rows and columns. For instance, if you have 5 rows with labels v, w, x, y, z

# slice from 'w' to 'y' and 'foo' to 'ant' by 3
df.loc['w':'y', 'foo':'ant':3]
#    foo ant
# w
# x
# y

Selecting columns by their position while using a function in pandas

Because working with Series, remove indexing columns:

def WoW(df):
     if df.iloc[1] == 0:
            return (df.iloc[1] - df.iloc[2])  
     else:
            return ((df.iloc[1] - df.iloc[2]) / df.iloc[1]) *100
frame['WoW%'] = frame.apply(WoW,axis=1)

Vectorized alternative:

s = frame.iloc[:,1] - frame.iloc[:,2]
frame['WoW%1'] = np.where(frame.iloc[:, 1] == 0, s, (s / frame.iloc[:,1]) *100)
print (frame)
   id  week1_values  week2_values        WoW%       WoW%1
0   1             0            32  -32.000000  -32.000000
1   2             0            35  -35.000000  -35.000000
2   3            13            25  -92.307692  -92.307692
3   4            39            78 -100.000000 -100.000000
4   5            64           200 -212.500000 -212.500000

pandas dataframe select columns in multiindex

There is a get_level_values method that you can use in conjunction with boolean indexing to get the the intended result.

In [13]:

df = pd.DataFrame(np.random.random((4,4)))
df.columns = pd.MultiIndex.from_product([[1,2],['A','B']])
print df
          1                   2          
          A         B         A         B
0  0.543980  0.628078  0.756941  0.698824
1  0.633005  0.089604  0.198510  0.783556
2  0.662391  0.541182  0.544060  0.059381
3  0.841242  0.634603  0.815334  0.848120
In [14]:

print df.iloc[:, df.columns.get_level_values(1)=='A']
          1         2
          A         A
0  0.543980  0.756941
1  0.633005  0.198510
2  0.662391  0.544060
3  0.841242  0.815334

Selecting/excluding sets of columns in pandas

You can either Drop the columns you do not need OR Select the ones you need

# Using DataFrame.drop
df.drop(df.columns[[1, 2]], axis=1, inplace=True)

# drop by Name
df1 = df1.drop(['B', 'C'], axis=1)

# Select the ones you want
df1 = df[['a','d']]

---

Related Topics