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.
10 Answers
7 Comments
tags
doesn't that remove the first 4 rows instead the first 3 rows in the original question?
bdiamante
No, It doesn't. The start position of the slice is always included.
citynorman
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:])
M.K
So if you want to delete from row 3 to row 9, for example, how would you do it?
df=df.iloc[3:9]?bdiamante
@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:]]). |
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)
7 Comments
tim
drop can even be calculated in-place (without extra assignment). Faster and simpler!
ChaimG
To expand on Tim's idea, Example:
df.drop(label, inplace=True)
Daniel Morgan
Due to index 0, I believe the implementation suggestion will delete 4 rows.
C Mars
@DanielMorgan That is not the case as python ranges are half open. As to why that is, is another question. See stackoverflow.com/questions/4504662/… or quora.com/…
|
df = df.iloc[n:]
n drops the first n rows.
Comments
A simple way is to use tail(-n) to remove the first n rows
df=df.tail(-3)
Comments
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.
Comments
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
2 Comments
cryanbhu
what does not in-place mean in
pandas?
Myles Hollowed
They're talking about the
inplace parameter that shows up in many methods in pandas.inp0= pd.read_csv("bank_marketing_updated_v1.csv",skiprows=2)
or if you want to do in existing dataframe
simply do following command
Comments
A bit late to answer, but I think the simplest way to remove the first 3 rows would just be:
df = df[3:]
Comments
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)
Comments
There is a simple way to implement that by the drop command.
df = df.drop(3)
2 Comments
fpersyn
Hi, Aman. Please wrap code examples in code blocks for clarity (see guidelines). It also helps to reference documentation pages to help the questioner understand the method.
Mustafa
this will fail if your index is not 1, 2, 3, 4, 5, for example.
header=3constructor argument which will set that row as the header row: stackoverflow.com/a/51822697/191246