Accessing Every 1St Element of Pandas Dataframe Column Containing Lists

Accessing every 1st element of Pandas DataFrame column containing lists

You can use map and a lambda function

df.loc[:, 'new_col'] = df.A.map(lambda x: x[0])

Getting the first element from each list in a column of lists

The easiest way is using str.get

# df['new_column'] = df['codes'].str.get(0)
df['new_column'] = df['codes'].str[0]

However, I would suggest a list comprehension for speed, if there are no NaNs:

df['new_column'] = [l[0] for l in df['codes']]

If lists can be empty, you can do something like:

df['new_column'] = [l[0] if len(l) > 0 else np.nan for l in df['codes']]

To handle NaNs with the list comprehension, you can use loc to subset and assign back.

m = df['codes'].notna()
df.loc[m, 'new_column'] = [
    l[0] if len(l) > 0 else np.nan for l in df.loc[m, 'codes']]

Obligatory why-is-list-comp-worth-it link: For loops with pandas - When should I care?

Accessing the first item of list values that occur irregularly in a Pandas column

You can remove '' from 'list' to list:

df1['Fruits'] = df1['Fruits'].apply(lambda  x : x[0] if type(x) == list else x)
print (df1)
   Fruits
0   Apple
1  Banana
2    Kiwi
3  Cheese

Extracting an element of a list in a pandas column

You can use the str accessor for lists, e.g.:

df_params['Gamma'].str[0]

This should work for all columns:

df_params.apply(lambda col: col.str[0])

Get first list element in apply function Pandas

Do you want:

df['runs'] = df['name'].apply(preprocess_names).str[0]

Python Pandas: selecting 1st element in array in all cells

You can use apply and for selecting first value of list use indexing with str:

print (zz.apply(lambda x: x.str[0]))
   0  1  2
0  1  2  3
1  4  1  4
2  1  2  3
3  4  1  4

Another solution with stack and unstack:

print (zz.stack().str[0].unstack())
   0  1  2
0  1  2  3
1  4  1  4
2  1  2  3
3  4  1  4

How to replace a list with first element of list in pandas dataframe column?

As you have strings, you could use a regex here:

df['Country'] = df['Country'].str.extract('((?<=\[["\'])[^"\']*|^[^"\']+$)')

output (as a new column for clarity):

     Name                 Country Country2
0   Harry                     USA      USA
1     Sam  ['USA', 'UK', 'India']      USA
2     Raj        ['India', 'USA']    India
3   Jamie                  Russia   Russia
4  Rupert                   China    China

regex:

(             # start capturing
(?<=\[["\'])  # if preceded by [" or ['
[^"\']*       # get all text until " or '
|             # OR
^[^"\']+$     # get whole string if it doesn't contain " or '
)             # stop capturing

Get first row value of a given column

To select the ith row, use iloc:

In [31]: df_test.iloc[0]
Out[31]: 
ATime     1.2
X         2.0
Y        15.0
Z         2.0
Btime     1.2
C        12.0
D        25.0
E        12.0
Name: 0, dtype: float64

To select the ith value in the Btime column you could use:

In [30]: df_test['Btime'].iloc[0]
Out[30]: 1.2

There is a difference between `df_test['Btime'].iloc[0]` (recommended) and `df_test.iloc[0]['Btime']`:

DataFrames store data in column-based blocks (where each block has a single
dtype). If you select by column first, a view can be returned (which is
quicker than returning a copy) and the original dtype is preserved. In contrast,
if you select by row first, and if the DataFrame has columns of different
dtypes, then Pandas copies the data into a new Series of object dtype. So
selecting columns is a bit faster than selecting rows. Thus, although
df_test.iloc[0]['Btime'] works, df_test['Btime'].iloc[0] is a little bit
more efficient.

There is a big difference between the two when it comes to assignment.
df_test['Btime'].iloc[0] = x affects df_test, but df_test.iloc[0]['Btime']
may not. See below for an explanation of why. Because a subtle difference in
the order of indexing makes a big difference in behavior, it is better to use single indexing assignment:

df.iloc[0, df.columns.get_loc('Btime')] = x

`df.iloc[0, df.columns.get_loc('Btime')] = x` (recommended):

The recommended way to assign new values to a
DataFrame is to avoid chained indexing, and instead use the method shown by
andrew,

df.loc[df.index[n], 'Btime'] = x

df.iloc[n, df.columns.get_loc('Btime')] = x

The latter method is a bit faster, because df.loc has to convert the row and column labels to
positional indices, so there is a little less conversion necessary if you use
df.iloc instead.

`df['Btime'].iloc[0] = x` works, but is not recommended:

Although this works, it is taking advantage of the way DataFrames are currently implemented. There is no guarantee that Pandas has to work this way in the future. In particular, it is taking advantage of the fact that (currently) df['Btime'] always returns a
view (not a copy) so df['Btime'].iloc[n] = x can be used to assign a new value
at the nth location of the Btime column of df.

Since Pandas makes no explicit guarantees about when indexers return a view versus a copy, assignments that use chained indexing generally always raise a SettingWithCopyWarning even though in this case the assignment succeeds in modifying df:

In [22]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])
In [24]: df['bar'] = 100
In [25]: df['bar'].iloc[0] = 99
/home/unutbu/data/binky/bin/ipython:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  self._setitem_with_indexer(indexer, value)

In [26]: df
Out[26]: 
  foo  bar
0   A   99  <-- assignment succeeded
2   B  100
1   C  100

`df.iloc[0]['Btime'] = x` does not work:

In contrast, assignment with df.iloc[0]['bar'] = 123 does not work because df.iloc[0] is returning a copy:

In [66]: df.iloc[0]['bar'] = 123
/home/unutbu/data/binky/bin/ipython:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy

In [67]: df
Out[67]: 
  foo  bar
0   A   99  <-- assignment failed
2   B  100
1   C  100

Warning: I had previously suggested df_test.ix[i, 'Btime']. But this is not guaranteed to give you the ith value since ix tries to index by label before trying to index by position. So if the DataFrame has an integer index which is not in sorted order starting at 0, then using ix[i] will return the row labeled i rather than the ith row. For example,

In [1]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])

In [2]: df
Out[2]: 
  foo
0   A
2   B
1   C

In [4]: df.ix[1, 'foo']
Out[4]: 'C'

Accessing Every 1St Element of Pandas Dataframe Column Containing Lists