CODEWITHSUNDEEP
pythonpandas
ropolo
ropolo

Reputation: 137

Moving desired row to the top of pandas Data Frame

In pandas, how can I copy or move a row to the top of the Data Frame without creating a copy of the Data Frame?

For example, I managed to do almost what I want with the code below, but I have the impression that there might be a better way to accomplish this:

import pandas as pd

df = pd.DataFrame({'Probe':['Test1','Test2','Test3'], 'Sequence':['AATGCGT','TGCGTAA','ATGCATG']})

df

   Probe Sequence
0  Test1  AATGCGT
1  Test2  TGCGTAA
2  Test3  ATGCATG

df_shifted = df.shift(1)

df_shifted

   Probe Sequence
0    NaN      NaN
1  Test1  AATGCGT
2  Test2  TGCGTAA


df_shifted.ix[0] = df.ix[2]

df_shifted

   Probe Sequence
0  Test3  ATGCATG
1  Test1  AATGCGT
2  Test2  TGCGTAA

Upvotes: 7

Views: 18787

Answers (4)

mbugert
mbugert

Reputation: 299

Here's an alternative that does not require a new column or sort_values.

With default index (RangeIndex):

df = pd.DataFrame({'Probe': ['Test1', 'Test2', 'Test3'],
                   'Sequence': ['AATGCGT', 'TGCGTAA', 'ATGCATG']})

   Probe Sequence
0  Test1  AATGCGT
1  Test2  TGCGTAA
2  Test3  ATGCATG

to_appear_first = [2]
new_index_order = [*to_appear_first, *df.index.difference(to_appear_first)]
df.loc[new_index_order].reset_index(drop=True)

   Probe Sequence
0  Test3  ATGCATG
1  Test1  AATGCGT
2  Test2  TGCGTAA

With Probe as the index column:

df = pd.DataFrame({'Probe': ['Test1', 'Test2', 'Test3'],
                   'Sequence': ['AATGCGT', 'TGCGTAA', 'ATGCATG']}).set_index("Probe")

      Sequence
Probe         
Test1  AATGCGT
Test2  TGCGTAA
Test3  ATGCATG

to_appear_first = ["Test3"]
new_index_order = [*to_appear_first, *df.index.difference(to_appear_first)]
df.loc[new_index_order]

      Sequence
Probe         
Test3  ATGCATG
Test1  AATGCGT
Test2  TGCGTAA

Upvotes: 0

Merlin
Merlin

Reputation: 25709

Try this. You don't need to make a copy of the dataframe.

df["new"] = range(1,len(df)+1)

   Probe Sequence  new
0  Test1  AATGCGT    1
1  Test2  TGCGTAA    2
2  Test3  ATGCATG    3


df.ix[2,'new'] = 0
df.sort_values("new").drop('new', axis=1)

   Probe Sequence
2  Test3  ATGCATG
0  Test1  AATGCGT
1  Test2  TGCGTAA

Basically, since you can't insert the row into the index at 0, create a column so you can.

If you want the index ordered, use this:

df.sort_values("new").reset_index(drop='True').drop('new', axis=1)

   Probe Sequence
0  Test3  ATGCATG
1  Test1  AATGCGT
2  Test2  TGCGTAA

Edit: df.ix is deprecated. Here's the same method with .loc.

df["new"] = range(1,len(df)+1)
df.loc[df.index==2, 'new'] = 0
df.sort_values("new").drop('new', axis=1)

Upvotes: 8

Kartik
Kartik

Reputation: 8703

pandas.concat:

df = pd.concat([df.iloc[[n],:], df.drop(n, axis=0)], axis=0)

Upvotes: 9

ropolo
ropolo

Reputation: 137

Okay, I think I came up with a solution. By all means, please feel free to add your own answer if you think yours is better:

import numpy as np

df.ix[3] = np.nan

df

   Probe Sequence
0  Test1  AATGCGT
1  Test2  TGCGTAA
2  Test3  ATGCATG
3    NaN      NaN

df = df.shift(1)

   Probe Sequence
0    NaN      NaN
1  Test1  AATGCGT
2  Test2  TGCGTAA
3  Test3  ATGCATG

df.ix[0] = df.ix[2]

df

   Probe Sequence
0  Test3  ATGCATG
1  Test1  AATGCGT
2  Test2  TGCGTAA
3  Test3  ATGCATG

Upvotes: 1

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