我看了其他答案,但看起来他们使用了2个不同的值。
代码:
user = ['X', 'Y', 'Z']
info = [['a','b','c',], ['d','e','f'], ['g','h','i']]
for u, g in user, range(len(user)):
print '|',u,'|',info[g][0],'|',info[g][1],'|',info[g][2],'| \n'
所以基本上,它需要输出:
'| X | a | b | c |'
'| Y | d | e | f |'
'| z | g | h | i |'
但相反,我得到了这个错误:
Traceback (most recent call last):
File "<pyshell#19>", line 1, in <module>
for u, g in user, range(len(user)):
ValueError: too many values to unpack
据我所知,用户和范围(len(用户))具有相同的价值。
答案 0 :(得分:14)
for u,g in user, range(len(user)):
实际上相当于:
for u,g in (user, range(len(user))):
即一个元组。它首先返回user
列表,然后返回range
。由于user
中的项目数量为3
,而LHS中您只有两个变量(u
,b
),因此您将收到该错误
>>> u,g = user
Traceback (most recent call last):
File "<ipython-input-162-022ea4a14c62>", line 1, in <module>
u,g = user
ValueError: too many values to unpack
你在这里寻找zip
(并使用字符串格式而不是手动连接):
>>> user = ['X', 'Y', 'Z']
>>> info = [['a','b','c',], ['d','e','f'], ['g','h','i']]
for u, g in zip(user, info):
print "| {} | {} |".format(u," | ".join(g))
...
| X | a | b | c |
| Y | d | e | f |
| Z | g | h | i |
答案 1 :(得分:2)
for u in user:
print '|',u,'|',info[user.index(u)][0],'|',info[user.index(u)][1],'|',info[user.index(u)][2],'| \n'
答案 2 :(得分:1)
>>> g = 0
>>> for u in user:
... print '|',u,'|',info[g][0],'|',info[g][1],'|',info[g][2],'| \n'
... g=g+1
...
| X | a | b | c |
| Y | d | e | f |
| Z | g | h | i |
答案 3 :(得分:0)
以下代码也有效:
>>> for i in zip(user,info):
print i[0],'|',
for x in i[1]:
print x,'|',
print
我得到了输出:
X | a | b | c |
Y | d | e | f |
Z | g | h | i |
P.S。:我知道有几个答案,但我只是认为这也是一种获得结果的方法,这就是我添加它的原因!