Get list from pandas dataframe column or row?

Question:

I have a dataframe df imported from an Excel document like this:

cluster load_date   budget  actual  fixed_price
A   1/1/2014    1000    4000    Y
A   2/1/2014    12000   10000   Y
A   3/1/2014    36000   2000    Y
B   4/1/2014    15000   10000   N
B   4/1/2014    12000   11500   N
B   4/1/2014    90000   11000   N
C   7/1/2014    22000   18000   N
C   8/1/2014    30000   28960   N
C   9/1/2014    53000   51200   N

I want to be able to return the contents of column 1 df['cluster'] as a list, so I can run a for-loop over it, and create an Excel worksheet for every cluster.

Is it also possible to return the contents of a whole column or row to a list? e.g.

list = [], list[column1] or list[df.ix(row1)]
Asked By: yoshiserry

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Answers:

Pandas DataFrame columns are Pandas Series when you pull them out, which you can then call x.tolist() on to turn them into a Python list. Alternatively you cast it with list(x).

import pandas as pd

data_dict = {'one': pd.Series([1, 2, 3], index=['a', 'b', 'c']),
             'two': pd.Series([1, 2, 3, 4], index=['a', 'b', 'c', 'd'])}

df = pd.DataFrame(data_dict)

print(f"DataFrame:n{df}n")
print(f"column types:n{df.dtypes}")

col_one_list = df['one'].tolist()

col_one_arr = df['one'].to_numpy()

print(f"ncol_one_list:n{col_one_list}ntype:{type(col_one_list)}")
print(f"ncol_one_arr:n{col_one_arr}ntype:{type(col_one_arr)}")

Output:

DataFrame:
   one  two
a  1.0    1
b  2.0    2
c  3.0    3
d  NaN    4

column types:
one    float64
two      int64
dtype: object

col_one_list:
[1.0, 2.0, 3.0, nan]
type:<class 'list'>

col_one_arr:
[ 1.  2.  3. nan]
type:<class 'numpy.ndarray'>
Answered By: Ben

This returns a numpy array:

arr = df["cluster"].to_numpy()

This returns a numpy array of unique values:

unique_arr = df["cluster"].unique()

You can also use numpy to get the unique values, although there are differences between the two methods:

arr = df["cluster"].to_numpy()
unique_arr = np.unique(arr)
Answered By: Anirudh Bandi

Example conversion:

Numpy Array -> Panda Data Frame -> List from one Panda Column

Numpy Array

data = np.array([[10,20,30], [20,30,60], [30,60,90]])

Convert numpy array into Panda data frame

dataPd = pd.DataFrame(data = data)
    
print(dataPd)
0   1   2
0  10  20  30
1  20  30  60
2  30  60  90

Convert one Panda column to list

pdToList = list(dataPd['2'])

Answered By: Hrvoje

Assuming the name of the dataframe after reading the excel sheet is df, take an empty list (e.g. dataList), iterate through the dataframe row by row and append to your empty list like-

dataList = [] #empty list
for index, row in df.iterrows(): 
    mylist = [row.cluster, row.load_date, row.budget, row.actual, row.fixed_price]
    dataList.append(mylist)

Or,

dataList = [] #empty list
for row in df.itertuples(): 
    mylist = [row.cluster, row.load_date, row.budget, row.actual, row.fixed_price]
    dataList.append(mylist)

No, if you print the dataList, you will get each rows as a list in the dataList.

Answered By: Natasha

As this question attained a lot of attention and there are several ways to fulfill your task, let me present several options.

Those are all one-liners by the way 😉

Starting with:

df
  cluster load_date budget actual fixed_price
0       A  1/1/2014   1000   4000           Y
1       A  2/1/2014  12000  10000           Y
2       A  3/1/2014  36000   2000           Y
3       B  4/1/2014  15000  10000           N
4       B  4/1/2014  12000  11500           N
5       B  4/1/2014  90000  11000           N
6       C  7/1/2014  22000  18000           N
7       C  8/1/2014  30000  28960           N
8       C  9/1/2014  53000  51200           N

Overview of potential operations:

ser_aggCol (collapse each column to a list)
cluster          [A, A, A, B, B, B, C, C, C]
load_date      [1/1/2014, 2/1/2014, 3/1/2...
budget         [1000, 12000, 36000, 15000...
actual         [4000, 10000, 2000, 10000,...
fixed_price      [Y, Y, Y, N, N, N, N, N, N]
dtype: object


ser_aggRows (collapse each row to a list)
0     [A, 1/1/2014, 1000, 4000, Y]
1    [A, 2/1/2014, 12000, 10000...
2    [A, 3/1/2014, 36000, 2000, Y]
3    [B, 4/1/2014, 15000, 10000...
4    [B, 4/1/2014, 12000, 11500...
5    [B, 4/1/2014, 90000, 11000...
6    [C, 7/1/2014, 22000, 18000...
7    [C, 8/1/2014, 30000, 28960...
8    [C, 9/1/2014, 53000, 51200...
dtype: object


df_gr (here you get lists for each cluster)
                             load_date                 budget                 actual fixed_price
cluster                                                                                         
A        [1/1/2014, 2/1/2014, 3/1/2...   [1000, 12000, 36000]    [4000, 10000, 2000]   [Y, Y, Y]
B        [4/1/2014, 4/1/2014, 4/1/2...  [15000, 12000, 90000]  [10000, 11500, 11000]   [N, N, N]
C        [7/1/2014, 8/1/2014, 9/1/2...  [22000, 30000, 53000]  [18000, 28960, 51200]   [N, N, N]


a list of separate dataframes for each cluster

df for cluster A
  cluster load_date budget actual fixed_price
0       A  1/1/2014   1000   4000           Y
1       A  2/1/2014  12000  10000           Y
2       A  3/1/2014  36000   2000           Y

df for cluster B
  cluster load_date budget actual fixed_price
3       B  4/1/2014  15000  10000           N
4       B  4/1/2014  12000  11500           N
5       B  4/1/2014  90000  11000           N

df for cluster C
  cluster load_date budget actual fixed_price
6       C  7/1/2014  22000  18000           N
7       C  8/1/2014  30000  28960           N
8       C  9/1/2014  53000  51200           N

just the values of column load_date
0    1/1/2014
1    2/1/2014
2    3/1/2014
3    4/1/2014
4    4/1/2014
5    4/1/2014
6    7/1/2014
7    8/1/2014
8    9/1/2014
Name: load_date, dtype: object


just the values of column number 2
0     1000
1    12000
2    36000
3    15000
4    12000
5    90000
6    22000
7    30000
8    53000
Name: budget, dtype: object


just the values of row number 7
cluster               C
load_date      8/1/2014
budget            30000
actual            28960
fixed_price           N
Name: 7, dtype: object


============================== JUST FOR COMPLETENESS ==============================


you can convert a series to a list
['C', '8/1/2014', '30000', '28960', 'N']
<class 'list'>


you can convert a dataframe to a nested list
[['A', '1/1/2014', '1000', '4000', 'Y'], ['A', '2/1/2014', '12000', '10000', 'Y'], ['A', '3/1/2014', '36000', '2000', 'Y'], ['B', '4/1/2014', '15000', '10000', 'N'], ['B', '4/1/2014', '12000', '11500', 'N'], ['B', '4/1/2014', '90000', '11000', 'N'], ['C', '7/1/2014', '22000', '18000', 'N'], ['C', '8/1/2014', '30000', '28960', 'N'], ['C', '9/1/2014', '53000', '51200', 'N']]
<class 'list'>

the content of a dataframe can be accessed as a numpy.ndarray
[['A' '1/1/2014' '1000' '4000' 'Y']
 ['A' '2/1/2014' '12000' '10000' 'Y']
 ['A' '3/1/2014' '36000' '2000' 'Y']
 ['B' '4/1/2014' '15000' '10000' 'N']
 ['B' '4/1/2014' '12000' '11500' 'N']
 ['B' '4/1/2014' '90000' '11000' 'N']
 ['C' '7/1/2014' '22000' '18000' 'N']
 ['C' '8/1/2014' '30000' '28960' 'N']
 ['C' '9/1/2014' '53000' '51200' 'N']]
<class 'numpy.ndarray'>

code:

# prefix ser refers to pd.Series object
# prefix df refers to pd.DataFrame object
# prefix lst refers to list object

import pandas as pd
import numpy as np

df=pd.DataFrame([
        ['A',   '1/1/2014',    '1000',    '4000',    'Y'],
        ['A',   '2/1/2014',    '12000',   '10000',   'Y'],
        ['A',   '3/1/2014',    '36000',   '2000',    'Y'],
        ['B',   '4/1/2014',    '15000',   '10000',   'N'],
        ['B',   '4/1/2014',    '12000',   '11500',   'N'],
        ['B',   '4/1/2014',    '90000',   '11000',   'N'],
        ['C',   '7/1/2014',    '22000',   '18000',   'N'],
        ['C',   '8/1/2014',    '30000',   '28960',   'N'],
        ['C',   '9/1/2014',    '53000',   '51200',   'N']
        ], columns=['cluster', 'load_date',   'budget',  'actual',  'fixed_price'])
print('df',df, sep='n', end='nn')

ser_aggCol=df.aggregate(lambda x: [x.tolist()], axis=0).map(lambda x:x[0])
print('ser_aggCol (collapse each column to a list)',ser_aggCol, sep='n', end='nnn')

ser_aggRows=pd.Series(df.values.tolist()) 
print('ser_aggRows (collapse each row to a list)',ser_aggRows, sep='n', end='nnn')

df_gr=df.groupby('cluster').agg(lambda x: list(x))
print('df_gr (here you get lists for each cluster)',df_gr, sep='n', end='nnn')

lst_dfFiltGr=[ df.loc[df['cluster']==val,:] for val in df['cluster'].unique() ]
print('a list of separate dataframes for each cluster', sep='n', end='nn')
for dfTmp in lst_dfFiltGr:
    print('df for cluster '+str(dfTmp.loc[dfTmp.index[0],'cluster']),dfTmp, sep='n', end='nn')

ser_singleColLD=df.loc[:,'load_date']
print('just the values of column load_date',ser_singleColLD, sep='n', end='nnn')

ser_singleCol2=df.iloc[:,2]
print('just the values of column number 2',ser_singleCol2, sep='n', end='nnn')

ser_singleRow7=df.iloc[7,:]
print('just the values of row number 7',ser_singleRow7, sep='n', end='nnn')

print('='*30+' JUST FOR COMPLETENESS '+'='*30, end='nnn')

lst_fromSer=ser_singleRow7.tolist()
print('you can convert a series to a list',lst_fromSer, type(lst_fromSer), sep='n', end='nnn')

lst_fromDf=df.values.tolist()
print('you can convert a dataframe to a nested list',lst_fromDf, type(lst_fromDf), sep='n', end='nn')

arr_fromDf=df.values
print('the content of a dataframe can be accessed as a numpy.ndarray',arr_fromDf, type(arr_fromDf), sep='n', end='nn')

as pointed out by cs95 other methods should be preferred over pandas .values attribute from pandas version 0.24 on see here. I use it here, because most people will (by 2019) still have an older version, which does not support the new recommendations. You can check your version with print(pd.__version__)

Answered By: Markus Dutschke
 amount = list()
    for col in df.columns:
        val = list(df[col])
        for v in val:
            amount.append(v)
Answered By: kamran kausar

If your column will only have one value something like pd.series.tolist() will produce an error. To guarantee that it will work for all cases, use the code below:

(
    df
        .filter(['column_name'])
        .values
        .reshape(1, -1)
        .ravel()
        .tolist()
)
Answered By: Ramin Melikov

If you do df.T.values.tolist() it generates list of lists of column values.

Answered By: Coddy

Here is simple one liner:

list(df['load_date'])

.toList() does not work anymore. It may have been the right API 10 years ago.

Answered By: David Dehghan

TL;DR: Use .tolist(). Don’t use list()

If we look at the source code of .tolist(), under the hood, list() function is being called on the underlying data in the dataframe, so both should produce the same output.

However, it looks like tolist() is optimized for columns of Python scalars because I found that calling list() on a column was 10 times slower than calling tolist(). For the record, I was trying to convert a column of json strings in a very large dataframe into a list and list() was taking its sweet time. That inspired me to test the runtimes of the two methods.


The following perfplot shows the runtime differences between the two methods on various pandas dtype Series objects. Basically, it’s showing the runtime differences between the following two methods:

list(df['some_col'])      # list()
df['some_col'].tolist()   # .tolist()

As you can see, no matter the size of the column/Series, for numeric and object dtype columns/Series, .tolist() method is much faster than list(). Not included here but the graphs for float and bool dtype columns were very similar to that of the int dtype column shown here. Also the graph for an object dtype column containing lists was very similar to the graph of string column shown here. Extension dtypes such as 'Int64Dtype', 'StringDtype', 'Float64Dtype' etc. also showed similar patterns.

On the other hand, there is virtually no difference between the two methods for datetime, timedelta and Categorical dtype columns.

perfplot

Code used to produce the above plot:

from perfplot import plot
kernels = [lambda s: list(s), lambda s: s.tolist()]
labels = ['list()', '.tolist()']
n_range = [2**k for k in range(4, 20)]
xlabel = 'Rows in DataFrame'
eq_chk = lambda x,y: all([x,y])

numeric = lambda n: pd.Series(range(5)).repeat(n)
string = lambda n: pd.Series(['some word', 'another word', 'a word']).repeat(n)
datetime = lambda n: pd.to_datetime(pd.Series(['2012-05-14', '2046-12-31'])).repeat(n)
timedelta = lambda n: pd.to_timedelta(pd.Series([1,2]), unit='D').repeat(n)
categorical = lambda n: pd.Series(pd.Categorical([1, 2, 3, 1, 2, 3])).repeat(n)

for n, f in [('Numeric', numeric), ('Object dtype', string), 
             ('Datetime', datetime), ('Timedelta', timedelta), 
             ('Categorical', categorical)]:
    plot(setup=f, kernels=kernels, labels=labels, n_range=n_range, 
         xlabel=xlabel, title=f'{n} column', equality_check=eq_chk);
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