The Python Oracle

Create a Pandas Dataframe by appending one row at a time

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Chapters
00:00 Question
00:36 Accepted answer (Score 874)
00:59 Answer 2 (Score 738)
01:44 Answer 3 (Score 405)
03:29 Answer 4 (Score 338)
03:55 Thank you

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Full question
https://stackoverflow.com/questions/1071...

Answer 2 links:
[NPE's answer]: https://stackoverflow.com/a/10716007/496...
[fred's answer]: https://stackoverflow.com/a/24888331/496...
[FooBar's answer]: https://stackoverflow.com/a/24913075/496...
[ShikharDua's answer]: https://stackoverflow.com/a/17496530/496...

Answer 3 links:
[Merge, join, and concatenate]: http://pandas.pydata.org/pandas-docs/sta...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #pandas #dataframe #append

#avk47



ACCEPTED ANSWER

Score 914


You can use df.loc[i], where the row with index i will be what you specify it to be in the dataframe.

>>> import pandas as pd
>>> from numpy.random import randint

>>> df = pd.DataFrame(columns=['lib', 'qty1', 'qty2'])
>>> for i in range(5):
>>>     df.loc[i] = ['name' + str(i)] + list(randint(10, size=2))

>>> df
     lib qty1 qty2
0  name0    3    3
1  name1    2    4
2  name2    2    8
3  name3    2    1
4  name4    9    6



ANSWER 2

Score 806


In case you can get all data for the data frame upfront, there is a much faster approach than appending to a data frame:

  1. Create a list of dictionaries in which each dictionary corresponds to an input data row.
  2. Create a data frame from this list.

I had a similar task for which appending to a data frame row by row took 30 min, and creating a data frame from a list of dictionaries completed within seconds.

rows_list = []
for row in input_rows:
    dict1 = {}
    # get input row in dictionary format
    # key = col_name
    dict1.update(blah..) 

    rows_list.append(dict1)

df = pd.DataFrame(rows_list)               



ANSWER 3

Score 457


In the case of adding a lot of rows to dataframe, I am interested in performance. So I tried the four most popular methods and checked their speed.

Performance

  1. Using .append (NPE's answer)
  2. Using .loc (fred's answer)
  3. Using .loc with preallocating (FooBar's answer)
  4. Using dict and create DataFrame in the end (ShikharDua's answer)

Runtime results (in seconds):

Approach 1000 rows 5000 rows 10 000 rows
.append 0.69 3.39 6.78
.loc without prealloc 0.74 3.90 8.35
.loc with prealloc 0.24 2.58 8.70
dict 0.012 0.046 0.084

So I use addition through the dictionary for myself.


Code:

import pandas as pd
import numpy as np
import time

del df1, df2, df3, df4
numOfRows = 1000
# append
startTime = time.perf_counter()
df1 = pd.DataFrame(np.random.randint(100, size=(5,5)), columns=['A', 'B', 'C', 'D', 'E'])
for i in range( 1,numOfRows-4):
    df1 = df1.append( dict( (a,np.random.randint(100)) for a in ['A','B','C','D','E']), ignore_index=True)
print('Elapsed time: {:6.3f} seconds for {:d} rows'.format(time.perf_counter() - startTime, numOfRows))
print(df1.shape)

# .loc w/o prealloc
startTime = time.perf_counter()
df2 = pd.DataFrame(np.random.randint(100, size=(5,5)), columns=['A', 'B', 'C', 'D', 'E'])
for i in range( 1,numOfRows):
    df2.loc[i]  = np.random.randint(100, size=(1,5))[0]
print('Elapsed time: {:6.3f} seconds for {:d} rows'.format(time.perf_counter() - startTime, numOfRows))
print(df2.shape)

# .loc with prealloc
df3 = pd.DataFrame(index=np.arange(0, numOfRows), columns=['A', 'B', 'C', 'D', 'E'] )
startTime = time.perf_counter()
for i in range( 1,numOfRows):
    df3.loc[i]  = np.random.randint(100, size=(1,5))[0]
print('Elapsed time: {:6.3f} seconds for {:d} rows'.format(time.perf_counter() - startTime, numOfRows))
print(df3.shape)

# dict
startTime = time.perf_counter()
row_list = []
for i in range (0,5):
    row_list.append(dict( (a,np.random.randint(100)) for a in ['A','B','C','D','E']))
for i in range( 1,numOfRows-4):
    dict1 = dict( (a,np.random.randint(100)) for a in ['A','B','C','D','E'])
    row_list.append(dict1)

df4 = pd.DataFrame(row_list, columns=['A','B','C','D','E'])
print('Elapsed time: {:6.3f} seconds for {:d} rows'.format(time.perf_counter() - startTime, numOfRows))
print(df4.shape)

P.S.: I believe my realization isn't perfect, and maybe there is some optimization that could be done.




ANSWER 4

Score 357


You could use pandas.concat(). For details and examples, see Merge, join, and concatenate.

For example:

def append_row(df, row):
    return pd.concat([
                df, 
                pd.DataFrame([row], columns=row.index)]
           ).reset_index(drop=True)

df = pd.DataFrame(columns=('lib', 'qty1', 'qty2'))
new_row = pd.Series({'lib':'A', 'qty1':1, 'qty2': 2})

df = append_row(df, new_row)