It’s good practice to always use the [] notation. One reason is that attribute notation (df.column_name) does not work for numbered indices:

In [1]: df = DataFrame([[1, 2, 3], [4, 5, 6]])

In [2]: df[1]
Out[2]:
0    2
1    5
Name: 1

In [3]: df.1
  File "<ipython-input-3-e4803c0d1066>", line 1
    df.1
       ^
SyntaxError: invalid syntax

The best way to do this in Pandas is to use drop:

df = df.drop('column_name', axis=1)

where 1 is the axis number (0 for rows and 1 for columns.)

Or, the drop() method accepts index/columns keywords as an alternative to specifying the axis. So we can now just do:

df = df.drop(columns=['column_nameA', 'column_nameB'])

• This was introduced in v0.21.0 (October 27, 2017)

To delete the column without having to reassign df you can do:

df.drop('column_name', axis=1, inplace=True)

Finally, to drop by column number instead of by column label, try this to delete, e.g. the 1st, 2nd and 4th columns:

df = df.drop(df.columns[[0, 1, 3]], axis=1)  # df.columns is zero-based pd.Index

Also working with "text" syntax for the columns:

df.drop(['column_nameA', 'column_nameB'], axis=1, inplace=True)

Use:

columns = ['Col1', 'Col2', ...]
df.drop(columns, inplace=True, axis=1)

This will delete one or more columns in-place. Note that inplace=True was added in pandas v0.13 and won’t work on older versions. You’d have to assign the result back in that case:

df = df.drop(columns, axis=1)

Drop by index

Delete first, second and fourth columns:

df.drop(df.columns[[0,1,3]], axis=1, inplace=True)

Delete first column:

df.drop(df.columns[[0]], axis=1, inplace=True)

There is an optional parameter inplace so that the original
data can be modified without creating a copy.

Popped

Column selection, addition, deletion

Delete column column-name:

df.pop('column-name')

Examples:

df = DataFrame.from_items([('A', [1, 2, 3]), ('B', [4, 5, 6]), ('C', [7,8, 9])], orient='index', columns=['one', 'two', 'three'])

print df:

   one  two  three
A    1    2      3
B    4    5      6
C    7    8      9

df.drop(df.columns[[0]], axis=1, inplace=True)
print df:

   two  three
A    2      3
B    5      6
C    8      9

three = df.pop('three')
print df:

   two
A    2
B    5
C    8

A nice addition is the ability to drop columns only if they exist. This way you can cover more use cases, and it will only drop the existing columns from the labels passed to it:

Simply add errors=’ignore’, for example.:

df.drop(['col_name_1', 'col_name_2', ..., 'col_name_N'], inplace=True, axis=1, errors='ignore')

• This is new from pandas 0.16.1 onward. Documentation is here.

In Pandas 0.16.1+, you can drop columns only if they exist per the solution posted by eiTan LaVi. Prior to that version, you can achieve the same result via a conditional list comprehension:

df.drop([col for col in ['col_name_1','col_name_2',...,'col_name_N'] if col in df],
        axis=1, inplace=True)

The dot syntax works in JavaScript, but not in Python.

• Python: del df['column_name']
• JavaScript: del df['column_name'] or del df.column_name

From version 0.16.1, you can do

df.drop(['column_name'], axis = 1, inplace = True, errors = 'ignore')

The actual question posed, missed by most answers here is:

Why can’t I use del df.column_name?

At first we need to understand the problem, which requires us to dive into Python magic methods.

As Wes points out in his answer, del df['column'] maps to the Python magic method df.__delitem__('column') which is implemented in Pandas to drop the column.

However, as pointed out in the link above about Python magic methods:

In fact, __del__ should almost never be used because of the precarious circumstances under which it is called; use it with caution!

You could argue that del df['column_name'] should not be used or encouraged, and thereby del df.column_name should not even be considered.

However, in theory, del df.column_name could be implemented to work in Pandas using the magic method __delattr__. This does however introduce certain problems, problems which the del df['column_name'] implementation already has, but to a lesser degree.

Example Problem

What if I define a column in a dataframe called "dtypes" or "columns"?

Then assume I want to delete these columns.

del df.dtypes would make the __delattr__ method confused as if it should delete the "dtypes" attribute or the "dtypes" column.

Architectural questions behind this problem

1. Is a dataframe a collection of columns?
2. Is a dataframe a collection of rows?
3. Is a column an attribute of a dataframe?

Pandas answers:

1. Yes, in all ways
2. No, but if you want it to be, you can use the .ix, .loc or .iloc methods.
3. Maybe, do you want to read data? Then yes, unless the name of the attribute is already taken by another attribute belonging to the dataframe. Do you want to modify data? Then no.

TLDR;

You cannot do del df.column_name, because Pandas has a quite wildly grown architecture that needs to be reconsidered in order for this kind of cognitive dissonance not to occur to its users.

Pro tip:

Don’t use df.column_name. It may be pretty, but it causes cognitive dissonance.

Zen of Python quotes that fits in here:

There are multiple ways of deleting a column.

There should be one– and preferably only one –obvious way to do it.

Columns are sometimes attributes but sometimes not.

Special cases aren’t special enough to break the rules.

Does del df.dtypes delete the dtypes attribute or the dtypes column?

In the face of ambiguity, refuse the temptation to guess.

TL;DR

A lot of effort to find a marginally more efficient solution. Difficult to justify the added complexity while sacrificing the simplicity of df.drop(dlst, 1, errors='ignore')

df.reindex_axis(np.setdiff1d(df.columns.values, dlst), 1)

Preamble
Deleting a column is semantically the same as selecting the other columns. I’ll show a few additional methods to consider.

I’ll also focus on the general solution of deleting multiple columns at once and allowing for the attempt to delete columns not present.

Using these solutions are general and will work for the simple case as well.

---

Setup
Consider the pd.DataFrame df and list to delete dlst

df = pd.DataFrame(dict(zip('ABCDEFGHIJ', range(1, 11))), range(3))
dlst = list('HIJKLM')

---

df

   A  B  C  D  E  F  G  H  I   J
0  1  2  3  4  5  6  7  8  9  10
1  1  2  3  4  5  6  7  8  9  10
2  1  2  3  4  5  6  7  8  9  10

---

dlst

['H', 'I', 'J', 'K', 'L', 'M']

The result should look like:

df.drop(dlst, 1, errors='ignore')

   A  B  C  D  E  F  G
0  1  2  3  4  5  6  7
1  1  2  3  4  5  6  7
2  1  2  3  4  5  6  7

---

Since I’m equating deleting a column to selecting the other columns, I’ll break it into two types:

1. Label selection
2. Boolean selection

---

Label Selection

We start by manufacturing the list/array of labels that represent the columns we want to keep and without the columns we want to delete.

1. df.columns.difference(dlst)

   Index(['A', 'B', 'C', 'D', 'E', 'F', 'G'], dtype='object')
   

2. np.setdiff1d(df.columns.values, dlst)

   array(['A', 'B', 'C', 'D', 'E', 'F', 'G'], dtype=object)
   

3. df.columns.drop(dlst, errors='ignore')

   Index(['A', 'B', 'C', 'D', 'E', 'F', 'G'], dtype='object')
   

4. list(set(df.columns.values.tolist()).difference(dlst))

   # does not preserve order
   ['E', 'D', 'B', 'F', 'G', 'A', 'C']
   

5. [x for x in df.columns.values.tolist() if x not in dlst]

   ['A', 'B', 'C', 'D', 'E', 'F', 'G']
   

   ---

Columns from Labels
For the sake of comparing the selection process, assume:

 cols = [x for x in df.columns.values.tolist() if x not in dlst]

Then we can evaluate

1. df.loc[:, cols]
2. df[cols]
3. df.reindex(columns=cols)
4. df.reindex_axis(cols, 1)

Which all evaluate to:

   A  B  C  D  E  F  G
0  1  2  3  4  5  6  7
1  1  2  3  4  5  6  7
2  1  2  3  4  5  6  7

---

Boolean Slice

We can construct an array/list of booleans for slicing

1. ~df.columns.isin(dlst)
2. ~np.in1d(df.columns.values, dlst)
3. [x not in dlst for x in df.columns.values.tolist()]
4. (df.columns.values[:, None] != dlst).all(1)

Columns from Boolean
For the sake of comparison

bools = [x not in dlst for x in df.columns.values.tolist()]

1. df.loc[: bools]

Which all evaluate to:

   A  B  C  D  E  F  G
0  1  2  3  4  5  6  7
1  1  2  3  4  5  6  7
2  1  2  3  4  5  6  7

---

Robust Timing

Functions

setdiff1d = lambda df, dlst: np.setdiff1d(df.columns.values, dlst)
difference = lambda df, dlst: df.columns.difference(dlst)
columndrop = lambda df, dlst: df.columns.drop(dlst, errors='ignore')
setdifflst = lambda df, dlst: list(set(df.columns.values.tolist()).difference(dlst))
comprehension = lambda df, dlst: [x for x in df.columns.values.tolist() if x not in dlst]

loc = lambda df, cols: df.loc[:, cols]
slc = lambda df, cols: df[cols]
ridx = lambda df, cols: df.reindex(columns=cols)
ridxa = lambda df, cols: df.reindex_axis(cols, 1)

isin = lambda df, dlst: ~df.columns.isin(dlst)
in1d = lambda df, dlst: ~np.in1d(df.columns.values, dlst)
comp = lambda df, dlst: [x not in dlst for x in df.columns.values.tolist()]
brod = lambda df, dlst: (df.columns.values[:, None] != dlst).all(1)

Testing

res1 = pd.DataFrame(
    index=pd.MultiIndex.from_product([
        'loc slc ridx ridxa'.split(),
        'setdiff1d difference columndrop setdifflst comprehension'.split(),
    ], names=['Select', 'Label']),
    columns=[10, 30, 100, 300, 1000],
    dtype=float
)

res2 = pd.DataFrame(
    index=pd.MultiIndex.from_product([
        'loc'.split(),
        'isin in1d comp brod'.split(),
    ], names=['Select', 'Label']),
    columns=[10, 30, 100, 300, 1000],
    dtype=float
)

res = res1.append(res2).sort_index()

dres = pd.Series(index=res.columns, name='drop')

for j in res.columns:
    dlst = list(range(j))
    cols = list(range(j // 2, j + j // 2))
    d = pd.DataFrame(1, range(10), cols)
    dres.at[j] = timeit('d.drop(dlst, 1, errors="ignore")', 'from __main__ import d, dlst', number=100)
    for s, l in res.index:
        stmt = '{}(d, {}(d, dlst))'.format(s, l)
        setp = 'from __main__ import d, dlst, {}, {}'.format(s, l)
        res.at[(s, l), j] = timeit(stmt, setp, number=100)

rs = res / dres

---

rs

                          10        30        100       300        1000
Select Label                                                           
loc    brod           0.747373  0.861979  0.891144  1.284235   3.872157
       columndrop     1.193983  1.292843  1.396841  1.484429   1.335733
       comp           0.802036  0.732326  1.149397  3.473283  25.565922
       comprehension  1.463503  1.568395  1.866441  4.421639  26.552276
       difference     1.413010  1.460863  1.587594  1.568571   1.569735
       in1d           0.818502  0.844374  0.994093  1.042360   1.076255
       isin           1.008874  0.879706  1.021712  1.001119   0.964327
       setdiff1d      1.352828  1.274061  1.483380  1.459986   1.466575
       setdifflst     1.233332  1.444521  1.714199  1.797241   1.876425
ridx   columndrop     0.903013  0.832814  0.949234  0.976366   0.982888
       comprehension  0.777445  0.827151  1.108028  3.473164  25.528879
       difference     1.086859  1.081396  1.293132  1.173044   1.237613
       setdiff1d      0.946009  0.873169  0.900185  0.908194   1.036124
       setdifflst     0.732964  0.823218  0.819748  0.990315   1.050910
ridxa  columndrop     0.835254  0.774701  0.907105  0.908006   0.932754
       comprehension  0.697749  0.762556  1.215225  3.510226  25.041832
       difference     1.055099  1.010208  1.122005  1.119575   1.383065
       setdiff1d      0.760716  0.725386  0.849949  0.879425   0.946460
       setdifflst     0.710008  0.668108  0.778060  0.871766   0.939537
slc    columndrop     1.268191  1.521264  2.646687  1.919423   1.981091
       comprehension  0.856893  0.870365  1.290730  3.564219  26.208937
       difference     1.470095  1.747211  2.886581  2.254690   2.050536
       setdiff1d      1.098427  1.133476  1.466029  2.045965   3.123452
       setdifflst     0.833700  0.846652  1.013061  1.110352   1.287831

---

fig, axes = plt.subplots(2, 2, figsize=(8, 6), sharey=True)
for i, (n, g) in enumerate([(n, g.xs(n)) for n, g in rs.groupby('Select')]):
    ax = axes[i // 2, i % 2]
    g.plot.bar(ax=ax, title=n)
    ax.legend_.remove()
fig.tight_layout()

This is relative to the time it takes to run df.drop(dlst, 1, errors='ignore'). It seems like after all that effort, we only improve performance modestly.

If fact the best solutions use reindex or reindex_axis on the hack list(set(df.columns.values.tolist()).difference(dlst)). A close second and still very marginally better than drop is np.setdiff1d.

rs.idxmin().pipe(
    lambda x: pd.DataFrame(
        dict(idx=x.values, val=rs.lookup(x.values, x.index)),
        x.index
    )
)

                      idx       val
10     (ridx, setdifflst)  0.653431
30    (ridxa, setdifflst)  0.746143
100   (ridxa, setdifflst)  0.816207
300    (ridx, setdifflst)  0.780157
1000  (ridxa, setdifflst)  0.861622

Pandas 0.21+ answer

Pandas version 0.21 has changed the drop method slightly to include both the index and columns parameters to match the signature of the rename and reindex methods.

df.drop(columns=['column_a', 'column_c'])

Personally, I prefer using the axis parameter to denote columns or index because it is the predominant keyword parameter used in nearly all pandas methods. But, now you have some added choices in version 0.21.

Another way of deleting a column in a Pandas DataFrame

If you’re not looking for in-place deletion then you can create a new DataFrame by specifying the columns using DataFrame(...) function as:

my_dict = { 'name' : ['a','b','c','d'], 'age' : [10,20,25,22], 'designation' : ['CEO', 'VP', 'MD', 'CEO']}

df = pd.DataFrame(my_dict)

Create a new DataFrame as

newdf = pd.DataFrame(df, columns=['name', 'age'])

You get a result as good as what you get with del / drop.

If your original dataframe df is not too big, you have no memory constraints, and you only need to keep a few columns, or, if you don’t know beforehand the names of all the extra columns that you do not need, then you might as well create a new dataframe with only the columns you need:

new_df = df[['spam', 'sausage']]

We can remove or delete a specified column or specified columns by the drop() method.

Suppose df is a dataframe.

Column to be removed = column0

Code:

df = df.drop(column0, axis=1)

To remove multiple columns col1, col2, . . . , coln, we have to insert all the columns that needed to be removed in a list. Then remove them by the drop() method.

Code:

df = df.drop([col1, col2, . . . , coln], axis=1)

Use:

df.drop('columnname', axis =1, inplace = True)

Or else you can go with

del df['colname']

To delete multiple columns based on column numbers

df.drop(df.iloc[:,1:3], axis = 1, inplace = True)

To delete multiple columns based on columns names

df.drop(['col1','col2',..'coln'], axis = 1, inplace = True)

Deleting a column using the iloc function of dataframe and slicing, when we have a typical column name with unwanted values:

df = df.iloc[:,1:] # Removing an unnamed index column

Here 0 is the default row and 1 is the first column, hence :,1: is our parameter for deleting the first column.

To remove columns before and after specific columns you can use the method truncate. For example:

   A   B    C     D      E
0  1  10  100  1000  10000
1  2  20  200  2000  20000

df.truncate(before='B', after='D', axis=1)

Output:

    B    C     D
0  10  100  1000
1  20  200  2000

Viewed from a general Python standpoint, del obj.column_name makes sense if the attribute column_name can be deleted. It needs to be a regular attribute – or a property with a defined deleter.

The reasons why this doesn’t translate to Pandas, and does not make sense for Pandas Dataframes are:

• Consider df.column_name to be a “virtual attribute”, it is not a thing in its own right, it is not the “seat” of that column, it’s just a way to access the column. Much like a property with no deleter.

Taking advantage by using Autocomplete or "IntelliSense" over string literals:

del df[df.column1.name]

# or

df.drop(df.column1.name, axis=1, inplace=True)

It works fine with current Pandas versions.

If you find the simple ways to delete column_name from df data frame, here we go:

df = df[df.columns.drop('column_name')]