问题：在满足熊猫特定条件的地方更新行值
说我有以下数据框：
什么是更新列的值最有效的方式壮举和another_feat其中流编号为2？
是这个吗？
for index, row in df.iterrows():
if df1.loc[index,'stream'] == 2:
# do something
更新： 如果我有超过100列怎么办？我不想明确命名要更新的列。我想将每列的值除以2（流列除外）。
所以要明确我的目标是：
将所有值除以具有流2的所有行的2，但不更改流列
Say I have the following dataframe:
What is the most efficient way to update the values of the columns feat and another_feat where the stream is number 2?
Is this it?
for index, row in df.iterrows():
if df1.loc[index,'stream'] == 2:
# do something
UPDATE: What to do if I have more than a 100 columns? I don’t want to explicitly name the columns that I want to update. I want to divide the value of each column by 2 (except for the stream column).
So to be clear what my goal is:
Dividing all values by 2 of all rows that have stream 2, but not changing the stream column
回答 0
我认为loc如果需要将两列更新为相同的值，可以使用：
df1.loc[df1['stream'] == 2, ['feat','another_feat']] = 'aaaa'
print df1
stream feat another_feat
a 1 some_value some_value
b 2 aaaa aaaa
c 2 aaaa aaaa
d 3 some_value some_value
如果需要单独更新，请使用以下一种方法：
df1.loc[df1['stream'] == 2, 'feat'] = 10
print df1
stream feat another_feat
a 1 some_value some_value
b 2 10 some_value
c 2 10 some_value
d 3 some_value some_value
另一个常见的选择是使用numpy.where：
df1['feat'] = np.where(df1['stream'] == 2, 10,20)
print df1
stream feat another_feat
a 1 20 some_value
b 2 10 some_value
c 2 10 some_value
d 3 20 some_value
编辑：如果您需要除stream条件之外的所有列True，请使用：
print df1
stream feat another_feat
a 1 4 5
b 2 4 5
c 2 2 9
d 3 1 7
#filter columns all without stream
cols = [col for col in df1.columns if col != 'stream']
print cols
['feat', 'another_feat']
df1.loc[df1['stream'] == 2, cols ] = df1 / 2
print df1
stream feat another_feat
a 1 4.0 5.0
b 2 2.0 2.5
c 2 1.0 4.5
d 3 1.0 7.0
回答 1
您可以使用进行相同的操作.ix，如下所示：
In [1]: df = pd.DataFrame(np.random.randn(5,4), columns=list('abcd'))
In [2]: df
Out[2]:
a b c d
0 -0.323772 0.839542 0.173414 -1.341793
1 -1.001287 0.676910 0.465536 0.229544
2 0.963484 -0.905302 -0.435821 1.934512
3 0.266113 -0.034305 -0.110272 -0.720599
4 -0.522134 -0.913792 1.862832 0.314315
In [3]: df.ix[df.a>0, ['b','c']] = 0
In [4]: df
Out[4]:
a b c d
0 -0.323772 0.839542 0.173414 -1.341793
1 -1.001287 0.676910 0.465536 0.229544
2 0.963484 0.000000 0.000000 1.934512
3 0.266113 0.000000 0.000000 -0.720599
4 -0.522134 -0.913792 1.862832 0.314315
编辑
在获得额外的信息之后，以下将返回所有列（满足某些条件的列）的值减半：
>> condition = df.a > 0
>> df[condition][[i for i in df.columns.values if i not in ['a']]].apply(lambda x: x/2)
我希望这有帮助！