Create a Pandas Dataframe by appending one row at a time

Question:

How do I create an empty DataFrame, then add rows, one by one?

I created an empty DataFrame:

df = pd.DataFrame(columns=('lib', 'qty1', 'qty2'))

Then I can add a new row at the end and fill a single field with:

df = df._set_value(index=len(df), col='qty1', value=10.0)

It works for only one field at a time. What is a better way to add new row to df?

Asked By: PhE

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Answers:

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

Answered By: NPE

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)               
Answered By: ShikharDua

For efficient appending, see How to add an extra row to a pandas dataframe and Setting With Enlargement.

Add rows through loc/ix on non existing key index data. For example:

In [1]: se = pd.Series([1,2,3])

In [2]: se
Out[2]:
0    1
1    2
2    3
dtype: int64

In [3]: se[5] = 5.

In [4]: se
Out[4]:
0    1.0
1    2.0
2    3.0
5    5.0
dtype: float64

Or:

In [1]: dfi = pd.DataFrame(np.arange(6).reshape(3,2),
   .....:                 columns=['A','B'])
   .....:

In [2]: dfi
Out[2]:
   A  B
0  0  1
1  2  3
2  4  5

In [3]: dfi.loc[:,'C'] = dfi.loc[:,'A']

In [4]: dfi
Out[4]:
   A  B  C
0  0  1  0
1  2  3  2
2  4  5  4
In [5]: dfi.loc[3] = 5

In [6]: dfi
Out[6]:
   A  B  C
0  0  1  0
1  2  3  2
2  4  5  4
3  5  5  5
Answered By: Nasser Al-Wohaibi

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
Answered By: fred

If you know the number of entries ex ante, you should preallocate the space by also providing the index (taking the data example from a different answer):

import pandas as pd
import numpy as np
# we know we're gonna have 5 rows of data
numberOfRows = 5
# create dataframe
df = pd.DataFrame(index=np.arange(0, numberOfRows), columns=('lib', 'qty1', 'qty2') )

# now fill it up row by row
for x in np.arange(0, numberOfRows):
    #loc or iloc both work here since the index is natural numbers
    df.loc[x] = [np.random.randint(-1,1) for n in range(3)]
In[23]: df
Out[23]: 
   lib  qty1  qty2
0   -1    -1    -1
1    0     0     0
2   -1     0    -1
3    0    -1     0
4   -1     0     0

Speed comparison

In[30]: %timeit tryThis() # function wrapper for this answer
In[31]: %timeit tryOther() # function wrapper without index (see, for example, @fred)
1000 loops, best of 3: 1.23 ms per loop
100 loops, best of 3: 2.31 ms per loop

And – as from the comments – with a size of 6000, the speed difference becomes even larger:

Increasing the size of the array (12) and the number of rows (500) makes
the speed difference more striking: 313ms vs 2.29s

Answered By: FooBar
mycolumns = ['A', 'B']
df = pd.DataFrame(columns=mycolumns)
rows = [[1,2],[3,4],[5,6]]
for row in rows:
    df.loc[len(df)] = row
Answered By: Lydia

You can append a single row as a dictionary using the ignore_index option.

>>> f = pandas.DataFrame(data = {'Animal':['cow','horse'], 'Color':['blue', 'red']})
>>> f
  Animal Color
0    cow  blue
1  horse   red
>>> f.append({'Animal':'mouse', 'Color':'black'}, ignore_index=True)
  Animal  Color
0    cow   blue
1  horse    red
2  mouse  black
Answered By: W.P. McNeill

This is not an answer to the OP question, but a toy example to illustrate ShikharDua’s answer which I found very useful.

While this fragment is trivial, in the actual data I had 1,000s of rows, and many columns, and I wished to be able to group by different columns and then perform the statistics below for more than one target column. So having a reliable method for building the data frame one row at a time was a great convenience. Thank you ShikharDua!

import pandas as pd

BaseData = pd.DataFrame({ 'Customer' : ['Acme','Mega','Acme','Acme','Mega','Acme'],
                          'Territory'  : ['West','East','South','West','East','South'],
                          'Product'  : ['Econ','Luxe','Econ','Std','Std','Econ']})
BaseData

columns = ['Customer','Num Unique Products', 'List Unique Products']

rows_list=[]
for name, group in BaseData.groupby('Customer'):
    RecordtoAdd={} #initialise an empty dict
    RecordtoAdd.update({'Customer' : name}) #
    RecordtoAdd.update({'Num Unique Products' : len(pd.unique(group['Product']))})
    RecordtoAdd.update({'List Unique Products' : pd.unique(group['Product'])})

    rows_list.append(RecordtoAdd)

AnalysedData = pd.DataFrame(rows_list)

print('Base Data : n',BaseData,'nn Analysed Data : n',AnalysedData)
Answered By: user3250815

Create a new record (data frame) and add to old_data_frame.

Pass a list of values and the corresponding column names to create a new_record (data_frame):

new_record = pd.DataFrame([[0, 'abcd', 0, 1, 123]], columns=['a', 'b', 'c', 'd', 'e'])

old_data_frame = pd.concat([old_data_frame, new_record])
Answered By: Jack Daniel

Another way to do it (probably not very performant):

# add a row
def add_row(df, row):
    colnames = list(df.columns)
    ncol = len(colnames)
    assert ncol == len(row), "Length of row must be the same as width of DataFrame: %s" % row
    return df.append(pd.DataFrame([row], columns=colnames))

You can also enhance the DataFrame class like this:

import pandas as pd
def add_row(self, row):
    self.loc[len(self.index)] = row
pd.DataFrame.add_row = add_row
Answered By: qed

For the sake of a Pythonic way:

res = pd.DataFrame(columns=('lib', 'qty1', 'qty2'))
res = res.append([{'qty1':10.0}], ignore_index=True)
print(res.head())

   lib  qty1  qty2
0  NaN  10.0   NaN
Answered By: hkyi

Make it simple. By taking a list as input which will be appended as a row in the data-frame:

import pandas as pd
res = pd.DataFrame(columns=('lib', 'qty1', 'qty2'))
for i in range(5):
    res_list = list(map(int, input().split()))
    res = res.append(pd.Series(res_list, index=['lib', 'qty1', 'qty2']), ignore_index=True)
Answered By: Vineet Jain

You can also build up a list of lists and convert it to a dataframe –

import pandas as pd

columns = ['i','double','square']
rows = []

for i in range(6):
    row = [i, i*2, i*i]
    rows.append(row)

df = pd.DataFrame(rows, columns=columns)

giving

    i   double  square
0   0   0   0
1   1   2   1
2   2   4   4
3   3   6   9
4   4   8   16
5   5   10  25
Answered By: Brian Burns

This will take care of adding an item to an empty DataFrame. The issue is that df.index.max() == nan for the first index:

df = pd.DataFrame(columns=['timeMS', 'accelX', 'accelY', 'accelZ', 'gyroX', 'gyroY', 'gyroZ'])

df.loc[0 if math.isnan(df.index.max()) else df.index.max() + 1] = [x for x in range(7)]
Answered By: tomatom

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.

Answered By: Mikhail_Sam

I figured out a simple and nice way:

>>> df
     A  B  C
one  1  2  3
>>> df.loc["two"] = [4,5,6]
>>> df
     A  B  C
one  1  2  3
two  4  5  6

Note the caveat with performance as noted in the comments.

Answered By: Qinsi

Here is the way to add/append a row in a Pandas DataFrame:

def add_row(df, row):
    df.loc[-1] = row
    df.index = df.index + 1
    return df.sort_index()

add_row(df, [1,2,3])

It can be used to insert/append a row in an empty or populated Pandas DataFrame.

Answered By: shivampip

We often see the construct df.loc[subscript] = … to assign to one DataFrame row. Mikhail_Sam posted benchmarks containing, among others, this construct as well as the method using dict and create DataFrame in the end. He found the latter to be the fastest by far.

But if we replace the df3.loc[i] = … (with preallocated DataFrame) in his code with df3.values[i] = …, the outcome changes significantly, in that that method performs similar to the one using dict. So we should more often take the use of df.values[subscript] = … into consideration. However note that .values takes a zero-based subscript, which may be different from the DataFrame.index.

Answered By: Armali

You can use a generator object to create a Dataframe, which will be more memory efficient over the list.

num = 10

# Generator function to generate generator object
def numgen_func(num):
    for i in range(num):
        yield ('name_{}'.format(i), (i*i), (i*i*i))

# Generator expression to generate generator object (Only once data get populated, can not be re used)
numgen_expression = (('name_{}'.format(i), (i*i), (i*i*i)) for i in range(num) )

df = pd.DataFrame(data=numgen_func(num), columns=('lib', 'qty1', 'qty2'))

To add raw to existing DataFrame you can use append method.

df = df.append([{ 'lib': "name_20", 'qty1': 20, 'qty2': 400  }])
Answered By: RockStar

Instead of a list of dictionaries as in ShikharDua’s answer, we can also represent our table as a dictionary of lists, where each list stores one column in row-order, given we know our columns beforehand. At the end we construct our DataFrame once.

For c columns and n rows, this uses one dictionary and c lists, versus one list and n dictionaries. The list-of-dictionaries method has each dictionary storing all keys and requires creating a new dictionary for every row. Here we only append to lists, which is constant time and theoretically very fast.

# Current data
data = {"Animal":["cow", "horse"], "Color":["blue", "red"]}

# Adding a new row (be careful to ensure every column gets another value)
data["Animal"].append("mouse")
data["Color"].append("black")

# At the end, construct our DataFrame
df = pd.DataFrame(data)
#   Animal  Color
# 0    cow   blue
# 1  horse    red
# 2  mouse  black
Answered By: qwr

pandas.DataFrame.append

DataFrame.append(self, other, ignore_index=False, verify_integrity=False, sort=False) → ‘DataFrame’

Code

df = pd.DataFrame([[1, 2], [3, 4]], columns=list('AB'))
df2 = pd.DataFrame([[5, 6], [7, 8]], columns=list('AB'))
df.append(df2)

With ignore_index set to True:

df.append(df2, ignore_index=True)
Answered By: kamran kausar

If you want to add a row at the end, append it as a list:

valuestoappend = [va1, val2, val3]
res = res.append(pd.Series(valuestoappend, index = ['lib', 'qty1', 'qty2']), ignore_index = True)
Answered By: Shahir Ansari

Before going to add a row, we have to convert the dataframe to a dictionary. There you can see the keys as columns in the dataframe and the values of the columns are again stored in the dictionary, but there the key for every column is the index number in the dataframe.

That idea makes me to write the below code.

df2 = df.to_dict()
values = ["s_101", "hyderabad", 10, 20, 16, 13, 15, 12, 12, 13, 25, 26, 25, 27, "good", "bad"] # This is the total row that we are going to add
i = 0
for x in df.columns:   # Here df.columns gives us the main dictionary key
    df2[x][101] = values[i]   # Here the 101 is our index number. It is also the key of the sub dictionary
    i += 1
Answered By: srikanth Gattu

You can concatenate two DataFrames for this. I basically came across this problem to add a new row to an existing DataFrame with a character index (not numeric).

So, I input the data for a new row in a duct() and index in a list.

new_dict = {put input for new row here}
new_list = [put your index here]

new_df = pd.DataFrame(data=new_dict, index=new_list)

df = pd.concat([existing_df, new_df])
Answered By: hansrajswapnil

All you need is loc[df.shape[0]] or loc[len(df)]


# Assuming your df has 4 columns (str, int, str, bool)
df.loc[df.shape[0]] = ['col1Value', 100, 'col3Value', False] 

or

df.loc[len(df)] = ['col1Value', 100, 'col3Value', False] 
Answered By: Giorgos Myrianthous
initial_data = {'lib': np.array([1,2,3,4]), 'qty1': [1,2,3,4], 'qty2': [1,2,3,4]}

df = pd.DataFrame(initial_data)

df

lib    qty1    qty2
0    1    1    1
1    2    2    2
2    3    3    3
3    4    4    4

val_1 = [10]
val_2 = [14]
val_3 = [20]

df.append(pd.DataFrame({'lib': val_1, 'qty1': val_2, 'qty2': val_3}))

lib    qty1    qty2
0    1    1    1
1    2    2    2
2    3    3    3
3    4    4    4
0    10    14    20

You can use a for loop to iterate through values or can add arrays of values.

val_1 = [10, 11, 12, 13]
val_2 = [14, 15, 16, 17]
val_3 = [20, 21, 22, 43]

df.append(pd.DataFrame({'lib': val_1, 'qty1': val_2, 'qty2': val_3}))

lib    qty1    qty2
0    1    1    1
1    2    2    2
2    3    3    3
3    4    4    4
0    10    14    20
1    11    15    21
2    12    16    22
3    13    17    43
Answered By: Harshal Deore

NEVER grow a DataFrame!

Yes, people have already explained that you should NEVER grow a DataFrame, and that you should append your data to a list and convert it to a DataFrame once at the end. But do you understand why?

Here are the most important reasons, taken from my post here.

  1. It is always cheaper/faster to append to a list and create a DataFrame in one go.
  2. Lists take up less memory and are a much lighter data structure to work with, append, and remove.
  3. dtypes are automatically inferred for your data. On the flip side, creating an empty frame of NaNs will automatically make them object, which is bad.
  4. An index is automatically created for you, instead of you having to take care to assign the correct index to the row you are appending.

This is The Right Way™ to accumulate your data

data = []
for a, b, c in some_function_that_yields_data():
    data.append([a, b, c])

df = pd.DataFrame(data, columns=['A', 'B', 'C'])

These options are horrible

  1. append or concat inside a loop

    append and concat aren’t inherently bad in isolation. The
    problem starts when you iteratively call them inside a loop – this
    results in quadratic memory usage.

    # Creates empty DataFrame and appends
    df = pd.DataFrame(columns=['A', 'B', 'C'])
    for a, b, c in some_function_that_yields_data():
        df = df.append({'A': i, 'B': b, 'C': c}, ignore_index=True)  
        # This is equally bad:
        # df = pd.concat(
        #       [df, pd.Series({'A': i, 'B': b, 'C': c})], 
        #       ignore_index=True)
    
  2. Empty DataFrame of NaNs

    Never create a DataFrame of NaNs as the columns are initialized with
    object (slow, un-vectorizable dtype).

    # Creates DataFrame of NaNs and overwrites values.
    df = pd.DataFrame(columns=['A', 'B', 'C'], index=range(5))
    for a, b, c in some_function_that_yields_data():
        df.loc[len(df)] = [a, b, c]
    

The Proof is in the Pudding

Timing these methods is the fastest way to see just how much they differ in terms of their memory and utility.

enter image description here

Benchmarking code for reference.


It’s posts like this that remind me why I’m a part of this community. People understand the importance of teaching folks getting the right answer with the right code, not the right answer with wrong code. Now you might argue that it is not an issue to use loc or append if you’re only adding a single row to your DataFrame. However, people often look to this question to add more than just one row – often the requirement is to iteratively add a row inside a loop using data that comes from a function (see related question). In that case it is important to understand that iteratively growing a DataFrame is not a good idea.

Answered By: cs95

If all data in your Dataframe has the same dtype you might use a NumPy array. You can write rows directly into the predefined array and convert it to a dataframe at the end.
It seems to be even faster than converting a list of dicts.

import pandas as pd
import numpy as np
from string import ascii_uppercase

startTime = time.perf_counter()
numcols, numrows = 5, 10000
npdf = np.ones((numrows, numcols))
for row in range(numrows):
    npdf[row, 0:] = np.random.randint(0, 100, (1, numcols))
df5 = pd.DataFrame(npdf, columns=list(ascii_uppercase[:numcols]))
print('Elapsed time: {:6.3f} seconds for {:d} rows'.format(time.perf_counter() - startTime, numOfRows))
print(df5.shape)
Answered By: Gerard

If you have a data frame df and want to add a list new_list as a new row to df, you can simply do:

df.loc[len(df)] = new_list

If you want to add a new data frame new_df under data frame df, then you can use:

df.append(new_df)
Answered By: Mahdi

If you always want to add a new row at the end, use this:

df.loc[len(df)] = ['name5', 9, 0]
Answered By: Prajot Kuvalekar

This code snippet uses a list of dictionaries to update the data frame. It adds on to ShikharDua’s and Mikhail_Sam’s answers.

import pandas as pd
colour = ["red", "big", "tasty"]
fruits = ["apple", "banana", "cherry"]
dict1={}
feat_list=[]
for x in colour:
    for y in fruits:
#         print(x, y)
        dict1 = dict([('x',x),('y',y)])
#         print(f'dict 1 {dict1}')
        feat_list.append(dict1)
#         print(f'feat_list {feat_list}')
feat_df=pd.DataFrame(feat_list)
feat_df.to_csv('feat1.csv')
Answered By: Joaquim
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