Pandas DataFrame column to list [duplicate]

问题

I am pulling a subset of data from a column based on conditions in another column being met.

I can get the correct values back but it is in pandas.core.frame.DataFrame. How do I convert that to list?

import pandas as pd

tst = pd.read_csv('C:\\SomeCSV.csv')

lookupValue = tst['SomeCol'] == "SomeValue"
ID = tst[lookupValue][['SomeCol']]
#How To convert ID to a list

回答1:

Use .values to get a numpy.array and then .tolist() to get a list.

For example:

import pandas as pd
df = pd.DataFrame({'a':[1,3,5,7,4,5,6,4,7,8,9],
                   'b':[3,5,6,2,4,6,7,8,7,8,9]})

Result:

>>> df['a'].values.tolist()
[1, 3, 5, 7, 4, 5, 6, 4, 7, 8, 9]

or you can just use

>>> df['a'].tolist()
[1, 3, 5, 7, 4, 5, 6, 4, 7, 8, 9]

To drop duplicates you can do one of the following:

>>> df['a'].drop_duplicates().values.tolist()
[1, 3, 5, 7, 4, 6, 8, 9]
>>> list(set(df['a'])) # as pointed out by EdChum
[1, 3, 4, 5, 6, 7, 8, 9]

回答2:

I'd like to clarify a few things:

As other answers have pointed out, the simplest thing to do is use pandas.Series.tolist(). I'm not sure why the top voted answer leads off with using pandas.Series.values.tolist() since as far as I can tell, it adds syntax/confusion with no added benefit.
tst[lookupValue][['SomeCol']] is a dataframe (as stated in the question), not a series (as stated in a comment to the question). This is because tst[lookupValue] is a dataframe, and slicing it with [['SomeCol']] asks for a list of columns (that list that happens to have a length of 1), resulting in a dataframe being returned. If you remove the extra set of brackets, as in tst[lookupValue]['SomeCol'], then you are asking for just that one column rather than a list of columns, and thus you get a series back.
You need a series to use pandas.Series.tolist(), so you should definitely skip the second set of brackets in this case. FYI, if you ever end up with a one-column dataframe that isn't easily avoidable like this, you can use pandas.DataFrame.squeeze() to convert it to a series.
tst[lookupValue]['SomeCol'] is getting a subset of a particular column via chained slicing. It slices once to get a dataframe with only certain rows left, and then it slices again to get a certain column. You can get away with it here since you are just reading, not writing, but the proper way to do it is tst.loc[lookupValue, 'SomeCol'] (which returns a series).
Using the syntax from #4, you could reasonably do everything in one line: ID = tst.loc[tst['SomeCol'] == 'SomeValue', 'SomeCol'].tolist()

Demo Code:

import pandas as pd
df = pd.DataFrame({'colA':[1,2,1],
                   'colB':[4,5,6]})
filter_value = 1

print "df"
print df
print type(df)

rows_to_keep = df['colA'] == filter_value
print "\ndf['colA'] == filter_value"
print rows_to_keep
print type(rows_to_keep)

result = df[rows_to_keep]['colB']
print "\ndf[rows_to_keep]['colB']"
print result
print type(result)

result = df[rows_to_keep][['colB']]
print "\ndf[rows_to_keep][['colB']]"
print result
print type(result)

result = df[rows_to_keep][['colB']].squeeze()
print "\ndf[rows_to_keep][['colB']].squeeze()"
print result
print type(result)

result = df.loc[rows_to_keep, 'colB']
print "\ndf.loc[rows_to_keep, 'colB']"
print result
print type(result)

result = df.loc[df['colA'] == filter_value, 'colB']
print "\ndf.loc[df['colA'] == filter_value, 'colB']"
print result
print type(result)

ID = df.loc[rows_to_keep, 'colB'].tolist()
print "\ndf.loc[rows_to_keep, 'colB'].tolist()"
print ID
print type(ID)

ID = df.loc[df['colA'] == filter_value, 'colB'].tolist()
print "\ndf.loc[df['colA'] == filter_value, 'colB'].tolist()"
print ID
print type(ID)

Result:

df
   colA  colB
0     1     4
1     2     5
2     1     6
<class 'pandas.core.frame.DataFrame'>

df['colA'] == filter_value
0     True
1    False
2     True
Name: colA, dtype: bool
<class 'pandas.core.series.Series'>

df[rows_to_keep]['colB']
0    4
2    6
Name: colB, dtype: int64
<class 'pandas.core.series.Series'>

df[rows_to_keep][['colB']]
   colB
0     4
2     6
<class 'pandas.core.frame.DataFrame'>

df[rows_to_keep][['colB']].squeeze()
0    4
2    6
Name: colB, dtype: int64
<class 'pandas.core.series.Series'>

df.loc[rows_to_keep, 'colB']
0    4
2    6
Name: colB, dtype: int64
<class 'pandas.core.series.Series'>

df.loc[df['colA'] == filter_value, 'colB']
0    4
2    6
Name: colB, dtype: int64
<class 'pandas.core.series.Series'>

df.loc[rows_to_keep, 'colB'].tolist()
[4, 6]
<type 'list'>

df.loc[df['colA'] == filter_value, 'colB'].tolist()
[4, 6]
<type 'list'>

回答3:

You can use pandas.Series.tolist

e.g.:

import pandas as pd
df = pd.DataFrame({'a':[1,2,3], 'b':[4,5,6]})

Run:

>>> df['a'].tolist()

You will get

>>> [1, 2, 3]

回答4:

The above solution is good if all the data is of same dtype. Numpy arrays are homogeneous containers. When you do df.values the output is an numpy array. So if the data has int and float in it then output will either have int or float and the columns will loose their original dtype. Consider df

a  b 
0  1  4
1  2  5 
2  3  6 

a    float64
b    int64

So if you want to keep original dtype, you can do something like

row_list = df.to_csv(None, header=False, index=False).split('\n')

this will return each row as a string.

['1.0,4', '2.0,5', '3.0,6', '']

Then split each row to get list of list. Each element after splitting is a unicode. We need to convert it required datatype.

def f(row_str): 
  row_list = row_str.split(',')
  return [float(row_list[0]), int(row_list[1])]

df_list_of_list = map(f, row_list[:-1])

[[1.0, 4], [2.0, 5], [3.0, 6]]

来源：https://stackoverflow.com/questions/23748995/pandas-dataframe-column-to-list

标签

python

pandas

tolist