By John

2012-11-16 06:26:40 8 Comments

When deleting a column in a DataFrame I use:

del df['column_name']

And this works great. Why can't I use the following?

del df.column_name

Since it is possible to access the column/Series as df.column_name, I expected this to work.


@LondonRob 2013-08-09 11:12:09

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

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

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

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)

Note: Introduced in v0.21.0 (October 27, 2017), the drop() method accepts index/columns keywords as an alternative to specifying the axis.

So we can now just do:

df.drop(columns=['B', 'C'])

@beardc 2013-12-10 20:13:32

Is this recommended over del for some reason?

@LondonRob 2013-12-11 12:20:39

@BirdJaguarIV I don't know of any performance improvement, but readability-wise, drop is a more SQL-like description of the operation in question. Couldn't del potentially be interpreted as setting all the values in that column to NaN?

@beardc 2013-12-11 19:58:09

I hadn't thought of reading it that way, but I guess I'm more used to pythonisms than I am to SQL. Maybe depends on who's going to be reading it? I'm also a fan of saving keystrokes when possible, all else equal :)

@Andy Hayden 2014-01-18 06:52:51

Does this make a copy though?

@Paul 2014-05-28 12:59:30

Though this method of deletion has its merits, this answer does not really answer the question being asked.

@LondonRob 2014-05-28 16:43:07

True @Paul, but due to the title of the question, most people arriving here will do so via trying to work out how to delete a column.

@hobs 2016-04-14 20:17:28

@beardc another advantage of drop over del is that drop allows you to drop multiple columns at once, perform the operation inplace or not, and also delete records along any axis (especially useful for a 3-D matrix or Panel)

@modulitos 2016-08-12 08:53:57

Another advantage of drop over del is that drop is part of the pandas API and contains documentation.

@user402516 2016-08-30 17:09:24

What does the axis parameter refer to?

@Clock Slave 2017-03-18 11:18:36

@user402516 the axis parameter specifies whether to look at row labels or column labels for deleting. 0 refers to rows, 1 being for columns

@ayhan 2017-10-02 12:42:34

Starting from version 0.21, drop accepts columns or rows parameters inline with the rename method:… if you want to update

@LondonRob 2017-10-02 13:27:18

@ayhan As far as I can see this is not released yet. But thanks for the heads-up. (Also: what a brilliant improvement in syntax! Thumbs up from me.)

@Guilherme Felipe Reis 2019-02-26 23:16:08

How to delete only if the column exists?, because if doesn't exist I get an error

@Deepak 2019-06-26 13:23:36

what does 'inplace = true', means here while deleting a column.

@pppery 2019-09-08 02:16:39

This doesn't answer the question, which is who del df.column_name doesn't work.

@Arthur D. Howland 2019-09-26 16:07:06

What if I wanted to drop columns 6-10? What does that look like?

@user6839822 2019-11-08 14:19:40

@beardc "fan of saving keystrokes" --> try Perl :)

@error2007s 2020-04-29 22:12:31

How make the drop conditional?

@billiam 2020-07-24 15:52:23

I believe that if you use the columns kwarg, e.g. df.drop(columns=['A', 'B']), then you don't have to specify axis=1.

@LondonRob 2020-07-27 10:28:41

@billiam if you double-check this to make sure, you can propose an edit to this answer!

@billiam 2020-07-27 14:39:40

@LondonRob Thank you. I did but it was rejected. Regards.

@billiam 2020-07-27 14:43:38

@LondonRob 2020-07-27 15:42:48

@billiam fixed. Thanks for the edit. It's clearly useful to know in this context.

@ccpizza 2020-03-15 17:57:51

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']]

@Littin Rajan 2020-04-19 13:58:43

We can Remove or Delete a specified column or sprcified columns by drop() method.

Suppose df is a dataframe.

Column to be removed = column0


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 drop() method.


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

I hope it would be helpful.

@RSM 2020-05-20 05:54:48

df = df.drop([col1, col2, . . . , coln], axis=1) this does not work if i specify a variable name in place of col1, col2 etc. I get error column not in axis when its definitely present. @Littin Could you help?

@firelynx 2016-05-03 09:48:51

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 implemeted 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 in 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.


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.


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.

@pppery 2018-02-22 19:27:53

"In fact, __del__ should almost never be used because of the precarious circumstances under which it is called; use it with caution!" is completely irrelevant here, as the method being used here is __delattr__.

@firelynx 2018-02-23 10:01:45

@ppperry you're miss-quoting. it's the del builtin that is meant, not the .__del__ instance method. The del builtin is mapping to __delattr__ and __delitem__ which is what I am building my argument on. So maybe you want to re-read what I wrote.

@pppery 2018-02-25 20:20:51

__ ... __ gets intrerpreted as bold markup by StackExchange

@cs95 2019-05-23 17:26:05

"Don't use df.column_name, It may be pretty, but it causes cognitive dissonance" What does this mean? I am not a psychologist so I have to look this up to understand what you mean. Also, quoting The Zen is meaningless because there are hundreds of valid ways to do the same thing in pandas.

@Wes McKinney 2012-11-21 03:12:31

As you've guessed, the right syntax is

del df['column_name']

It's difficult to make del df.column_name work simply as the result of syntactic limitations in Python. del df[name] gets translated to df.__delitem__(name) under the covers by Python.

@dwanderson 2016-10-04 14:24:22

I realize this is a super old "answer", but my curiosity is piqued - why is that a syntactic limitation of Python? class A(object): def __init__(self): self.var = 1 sets up a class, then a = A(); del a.var works just fine...

@Yonatan 2016-12-22 08:27:07

@dwanderson the difference is that when a column is to be removed, the DataFrame needs to have its own handling for "how to do it". In the case of del df[name], it gets translated to df.__delitem__(name) which is a method that DataFrame can implement and modify to its needs. In the case of del, the member variable gets removed without a chance for any custom-code running. Consider your own example - can you get del a.var to result in a print of "deleting variable"? If you can, please tell me how. I can't :)

@Eugene Pakhomov 2017-01-19 16:06:35

@Yonatan 2017-01-22 19:03:27

@EugenePakhomov good point. I was answering in python 2, indeed python 3 gives more flexibility in such matters. Thanks for clarifying.

@C S 2017-06-20 12:38:31

@Yonatan Eugene's comment applies to Python 2 also; descriptors have been in Python 2 since 2.2 and it is trivial to satisfy your requirement ;)

@wizzwizz4 2017-09-30 09:42:52

This answer isn't really correct - the pandas developers didn't, but that doesn't mean it is hard to do.

@pedjjj 2020-03-30 17:04:50

Still, it is correct given that panda works / used to work on Py2.7, too - where you can't

@Zeynab Rostami 2020-09-13 12:17:09

using this answer may cause tokenizing problems when you save the csv and want to read it again. using "df.drop()" as @LondonRob described is the correct way.

@Ted Petrou 2017-10-24 14:31:03

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.

@YouAreAwesome 2018-04-22 05:03:59

df.drop(['column_a', 'column_c'], axis=1) | it is working for me for now

@Daksh 2018-09-09 06:59:06

Another way of Deleting a Column in 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

@cs95 2019-05-23 17:24:28

This is technically correct but it seems silly to have to list every column to keep instead of just the one (or few) columns you want to delete.

@Doctor 2016-04-20 15:55:38

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

@eiTan LaVi 2016-01-03 12:29:49

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.

@jezrael 2015-07-15 13:37:23

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.


Column selection, addition, deletion

Delete column column-name:



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:

A    2
B    5
C    8

@Kennet Celeste 2017-02-09 16:10:36

How can I pop a row in pandas?

@Clock Slave 2017-03-18 11:21:48

@Yugi You can use a transposed dataframe for that. ex - df.T.pop('A')

@cs95 2019-05-23 17:28:44

@ClockSlave That doesn't modify the original df. You could do df = df.T; df.pop(index); df = df.T but this seems excessive.

@Anirban Mukherjee 2019-12-04 20:02:28

Instead of df.drop(df.columns[[0]], axis=1, inplace=True) wouldn't it be enough to use df.drop([0], axis=1) ?

@jezrael 2019-12-04 20:42:43

@Anirban Mukherjee It depends. If want delete column name 0, then df.drop(0, axis=1) working well. But if dont know column name and need remove first column then need df.drop(df.columns[[0]], axis=1, inplace=True), it select first column by position and drop it.

@marsipan 2020-04-30 19:56:06

@KennetCeleste to pop rows: df.drop(df.index[[1,2]], axis=0, inplace=True) will pop rows 1 and 2.

@Tiago Martins Peres 李大仁 2020-06-14 08:02:07

In my case had to remove inplace=True for it to work.

@Krishna Sankar 2014-03-23 20:57:57


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)

@edesz 2017-06-14 23:31:19

A note about this answer: if a 'list' is used, the square brackets should be dropped: df.drop(list,inplace=True,axis=1)

@dbliss 2017-07-04 21:27:59

this should really be the accepted answer, because it makes clear the superiority of this method over del -- can drop more than one column at once.

@billiam 2020-07-24 15:53:54

I believe that if you use the columns kwarg, e.g. df.drop(columns=['A', 'B']), then you don't have to specify axis=1.

@micstr 2020-09-10 09:34:58

Latecomers also look below to @eiTanLaVi.solution for pandas 0.16.1+ who recommends add errors='ignore'

@Andy Hayden 2012-11-16 11:33:47

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]
0    2
1    5
Name: 1

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

@piRSquared 2017-09-20 05:43:19


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)

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.

Consider the pd.DataFrame df and list to delete dlst

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


   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


['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


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)


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

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

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)[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)[(s, l), j] = timeit(stmt, setp, number=100)

rs = res / dres


                          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], title=n)

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.

enter image description here

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.

    lambda x: pd.DataFrame(
        dict(idx=x.values, val=rs.lookup(x.values, 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

@Alexander 2016-02-13 21:58:33

In pandas 0.16.1+ you can drop columns only if they exist per the solution posted by @eiTanLaVi. 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)

@sushmit 2016-04-30 18:57:48

from version 0.16.1 you can do

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

@muon 2016-10-21 19:57:19

And this also supports dropping multiple columns, some of which need not exist (i.e. without raising error errors= 'ignore') df.drop(['column_1','column_2'], axis=1 , inplace=True,errors= 'ignore'), if such an application desired!

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