26 
In [35]: test = pd.DataFrame({'a':range(4),'b':range(4,8)})

In [36]: test
Out[36]: 
   a  b
0  0  4
1  1  5
2  2  6
3  3  7

In [37]: for i in test['a']:
   ....:  print i
   ....: 
0
1
2
3

In [38]: for i,j in test:
   ....:  print i,j
   ....: 
------------------------------------------------------------
Traceback (most recent call last):
  File "<ipython console>", line 1, in <module>
ValueError: need more than 1 value to unpack


In [39]: for i,j in test[['a','b']]:
   ....:  print i,j
   ....: 
------------------------------------------------------------
Traceback (most recent call last):
  File "<ipython console>", line 1, in <module>
ValueError: need more than 1 value to unpack


In [40]: for i,j in [test['a'],test['b']]:
   ....:  print i,j
   ....: 
------------------------------------------------------------
Traceback (most recent call last):
  File "<ipython console>", line 1, in <module>
ValueError: too many values to unpack
4

4 回答 4

43

使用DataFrame.itertuples()方法：

for a, b in test.itertuples(index=False):
    print a, b
于 2013-02-28T00:56:25.650 回答
23

您可以使用zip（这在 python 3 中是本机的，可以从python 2.7 中导入）itertools：izip

蟒蛇 3

for a,b in zip(test.a, test.b): 
    print(a,b)

蟒蛇 2

for a,b in izip(test.a, test.b): 
    print a,b
于 2016-09-09T20:47:55.753 回答
6

尝试，

for i in test.index : print test['a'][i], test['b'][i]

为你带来，

0 4
1 5
2 6
3 7
于 2013-02-28T00:45:11.420 回答
5

我仍在尝试自己学习熊猫。

您也可以使用该.iterrows()方法，它返回Index和Series每行：

test = DataFrame({'a':range(4),'b':range(4,8)})
for idx, series in test.iterrows():
    print series['a'], series['b']
于 2013-02-28T00:57:39.527 回答