This seems like a ridiculously easy question... but I'm not seeing the easy answer I was expecting.
So, how do I get the value at an nth row of a given column in Pandas? (I am particularly interested in the first row, but would be interested in a more general practice as well).
For example, let's say I want to pull the 1.2 value in Btime
as a variable.
Whats the right way to do this?
>>> df_test
ATime X Y Z Btime C D E
0 1.2 2 15 2 1.2 12 25 12
1 1.4 3 12 1 1.3 13 22 11
2 1.5 1 10 6 1.4 11 20 16
3 1.6 2 9 10 1.7 12 29 12
4 1.9 1 1 9 1.9 11 21 19
5 2.0 0 0 0 2.0 8 10 11
6 2.4 0 0 0 2.4 10 12 15
ベストアンサー1
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
or
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
警告: 以前 を提案しました。しかし、位置でインデックス付けする前にラベルでインデックス付けしようとするため、値df_test.ix[i, 'Btime']
が返される保証はありません。したがって、DataFrame に 0 から始まるソート順ではない整数インデックスがある場合、 を使用すると行ではなくラベルの付いた行が返されます。たとえば、ith
ix
ix[i]
i
ith
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'