Delete the first three rows of a dataframe in pandas

Question

I need to delete the first three rows of a dataframe in pandas.

I know df.ix[:-1] would remove the last row, but I can't figure out how to remove first n rows.

For read_csv/read_html you can use e.g. `header=3` constructor argument which will set that row as the header row: https://stackoverflow.com/a/51822697/191246 — ccpizza, Dec 16 '19 at 17:04

score 437 · Accepted Answer · edited Oct 17 '18 at 01:06

437

Use iloc:

df = df.iloc[3:]

will give you a new df without the first three rows.

edited Oct 17 '18 at 01:06

Asclepius

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answered May 06 '13 at 12:04

bdiamante

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doesn't that remove the first 4 rows instead the first 3 rows in the original question? – tagoma Nov 17 '16 at 18:16
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No, It doesn't. The start position of the slice is always included. – bdiamante Jan 20 '17 at 19:30
Anyone happen to know how to do this in a `groupby()`? This works but returns duplicate columns in the index `df=pd.DataFrame({'v':np.arange(10).tolist()*2,'g':['a']*10+['b']*10});df.groupby('g').apply(lambda x: x.iloc[3:])` – citynorman Aug 06 '17 at 22:24
So if you want to delete from row 3 to row 9, for example, how would you do it? `df=df.iloc[3:9]`? – M.K Jun 26 '19 at 14:25
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@M.K if using this approach, you can use this in combination with `pd.concat()`. Something like, `df2 = pd.concat([df.iloc[:3],df.iloc[10:]])`. – bdiamante Jun 26 '19 at 17:00
If you do this first row will be 3, how can I make it so it starts from 0. – haneulkim Jul 10 '19 at 04:24
@makewhite you can use `df = df.iloc[3:].reset_index(drop=True)` to reset the index to 0. – bdiamante Jul 10 '19 at 15:58

C Mars · Answer 2 · 2016-03-08T09:14:59.647

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I think a more explicit way of doing this is to use drop.

The syntax is:

df.drop(label)

And as pointed out by @tim and @ChaimG, this can be done in-place:

df.drop(label, inplace=True)

One way of implementing this could be:

df.drop(df.index[:3], inplace=True)

And another "in place" use:

df.drop(df.head(3).index, inplace=True)

edited Mar 08 '16 at 09:14

answered Jul 30 '13 at 14:39

C Mars

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`drop` can even be calculated in-place (without extra assignment). Faster and simpler! – tim Jun 13 '14 at 18:24
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To expand on Tim's idea, Example: `df.drop(label, inplace=True)` – ChaimG Aug 26 '15 at 04:41
Due to index 0, I believe the implementation suggestion will delete 4 rows. – Daniel Morgan May 07 '16 at 18:34
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@DanielMorgan That is not the case as python ranges are half open. As to why that is, is another question. See http://stackoverflow.com/questions/4504662/why-does-rangestart-end-not-include-end or https://www.quora.com/Why-are-Python-ranges-half-open-exclusive-instead-of-closed-inclusive – C Mars May 09 '16 at 07:10
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@tim, according to [this](https://stackoverflow.com/questions/22532302/pandas-peculiar-performance-drop-for-inplace-rename-after-dropna/22533110#22533110), `inplace` operations are not faster. Also, simpler is a matter of opinion: I find it easier to read when the code doesn't have `inplace` parameters. – toto_tico Jul 26 '17 at 09:14
@toto_tico Afaik, inplace operations are ideally intended to avoid duplication in memory. That can be a big deal for a large object. Even if an old implementation uses a copy, an ideal implementation can avoid a copy. – Asclepius Oct 17 '18 at 01:04

score 16 · Answer 3 · edited Oct 17 '18 at 01:02

16

df = df.iloc[n:]

n drops the first n rows.

edited Oct 17 '18 at 01:02

Asclepius

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answered May 16 '17 at 10:50

176coding

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score 12 · Answer 4 · answered Oct 25 '18 at 07:49

12

A simple way is to use tail(-n) to remove the first n rows

df=df.tail(-3)

answered Oct 25 '18 at 07:49

mxia

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score 8 · Answer 5 · edited Jul 20 '17 at 00:33

8

df.drop(df.index[[0,2]])

Pandas uses zero based numbering, so 0 is the first row, 1 is the second row and 2 is the third row.

edited Jul 20 '17 at 00:33

Steffen Winkler

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answered Jul 19 '17 at 19:39

Anupam khare

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score 7 · Answer 6 · answered May 10 '13 at 17:25

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You can use python slicing, but note it's not in-place.

In [15]: import pandas as pd
In [16]: import numpy as np
In [17]: df = pd.DataFrame(np.random.random((5,2)))
In [18]: df
Out[18]:
          0         1
0  0.294077  0.229471
1  0.949007  0.790340
2  0.039961  0.720277
3  0.401468  0.803777
4  0.539951  0.763267

In [19]: df[3:]
Out[19]:
          0         1
3  0.401468  0.803777
4  0.539951  0.763267

answered May 10 '13 at 17:25

beardc

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what does not in-place mean in `pandas`? – cryanbhu Oct 09 '19 at 08:20
They're talking about the `inplace` parameter that shows up in many methods in pandas. – Myles Hollowed Apr 07 '21 at 17:54

score 5 · Answer 7 · answered May 21 '20 at 18:40

5

inp0= pd.read_csv("bank_marketing_updated_v1.csv",skiprows=2)

or if you want to do in existing dataframe

simply do following command

answered May 21 '20 at 18:40

Rahul kuchhadia

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score 0 · Answer 8 · answered Apr 15 '23 at 14:16

Flexible truncation in pandas with `truncate`

To remove the first N rows

df.truncate(before=N)

To remove the last M rows

df.truncate(after=M)

To remove the first N and last M rows together

df.truncate(before=N, after=M)

To remove the first N columns

df.truncate(before=N, axis=1)

score -1 · Answer 9 · edited Aug 24 '21 at 18:34

-1

There is a simple way to implement that by the drop command.

df = df.drop(3)

edited Aug 24 '21 at 18:34

Flair

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answered Aug 22 '21 at 20:05

Aman Babbar

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Hi, Aman. Please wrap code examples in code blocks for clarity ([see guidelines](https://stackoverflow.com/editing-help#code)). It also helps to reference documentation pages to help the questioner understand the method. – fpersyn Aug 23 '21 at 16:01
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this will fail if your index is not 1, 2, 3, 4, 5, for example. – Mustafa Mar 19 '22 at 02:05

Delete the first three rows of a dataframe in pandas

9 Answers9

Flexible truncation in pandas with `truncate`

Linked

Delete the first three rows of a dataframe in pandas

9 Answers9

Flexible truncation in pandas with truncate

Linked

Flexible truncation in pandas with `truncate`