Create a Pandas Dataframe by appending one row at a time
PythonPandasDataframeAppendPython Problem Overview
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
?
Python Solutions
Solution 1 - Python
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
Solution 2 - Python
In case you can get all data for the data frame upfront, there is a much faster approach than appending to a data frame:
- Create a list of dictionaries in which each dictionary corresponds to an input data row.
- 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)
Solution 3 - Python
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
- Using .append (NPE's answer)
- Using .loc (fred's answer)
- Using .loc with preallocating (FooBar's answer)
- 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.
Solution 4 - Python
You could use pandas.concat()
or DataFrame.append()
. For details and examples, see Merge, join, and concatenate.
Solution 5 - Python
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.
- It is always cheaper/faster to append to a list and create a DataFrame in one go.
- Lists take up less memory and are a much lighter data structure to work with, append, and remove.
dtypes
are automatically inferred for your data. On the flip side, creating an empty frame of NaNs will automatically make themobject
, which is bad.- 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
-
append
orconcat
inside a loopappend
andconcat
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)
-
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.
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.
Solution 6 - Python
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
Solution 7 - Python
mycolumns = ['A', 'B']
df = pd.DataFrame(columns=mycolumns)
rows = [[1,2],[3,4],[5,6]]
for row in rows:
df.loc[len(df)] = row
Solution 8 - Python
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
Solution 9 - Python
For efficient appending, see https://stackoverflow.com/questions/19365513/how-to-add-an-extra-row-to-a-pandas-dataframe/19368360#19368360 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
Solution 10 - Python
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
Solution 11 - Python
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
Solution 12 - Python
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.
Solution 13 - Python
If you always want to add a new row at the end, use this:
df.loc[len(df)] = ['name5', 9, 0]
Solution 14 - Python
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,'\n\n Analysed Data : \n',AnalysedData)
Solution 15 - Python
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 }])
Solution 16 - Python
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])
Solution 17 - Python
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
Solution 18 - Python
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.
Solution 19 - Python
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)
Solution 20 - Python
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
Solution 21 - Python
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]
Solution 22 - Python
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
Solution 23 - Python
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])
Solution 24 - Python
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)
Solution 25 - Python
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.
Solution 26 - Python
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)
Solution 27 - Python
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
Solution 28 - Python
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)
Solution 29 - Python
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)
Solution 30 - Python
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')
Solution 31 - Python
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)]