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
答案 0 :(得分:21)
使用DataFrame.itertuples()
方法:
for a, b in test.itertuples(index=False):
print a, b
答案 1 :(得分:9)
你可以使用zip
(这在python 3中是原生的,可以在{2.7}中从itertools
导入为izip
:
for a,b in zip(test.a, test.b):
print(a,b)
for a,b in izip(test.a, test.b):
print a,b
答案 2 :(得分:3)
尝试,
for i in test.index : print test['a'][i], test['b'][i]
给你,
0 4
1 5
2 6
3 7
答案 3 :(得分:2)
我自己还在努力学习大熊猫。
您还可以使用.iterrows()
方法,每行返回Index
和Series
:
test = DataFrame({'a':range(4),'b':range(4,8)})
for idx, series in test.iterrows():
print series['a'], series['b']