如果python中有任何其他元素匹配，则打印元组的第一个元素

Question

I have a tuple in a list like

A = [(2, 'two', '2nd', 'second')]

and I am getting a variable 'B' which is possibly rest of elements in the list 'A' except first element.

For example:

B = ['two'] # or ['2nd'], or ['second']

I want to print first element of tuple 'A' if the variable 'B' matches with tuple 'A'

I have tried

[x for x,y,z,t in a[0] if  b[0] == y or b[0] == z or b[0] == t ]

I am getting below error.

TypeError: 'int' object is not iterable

Please help.

Answer 1

try this

li = []
for b in B:
    li += [a[0] for a in A if b in a[1:]]

This will work for n number of items in A and m number of items in B. Final output will be the first element of tuples in the list A which have at least one value that matches any element of B

Hope this will solve your issue

UPDATE

Shorthand will be

[a[0] for b in B for a in A if b in a[1:]]

Forgot to add in the first place.

Sample output

Python 2.7.12 (default, Nov 19 2016, 06:48:10) 
[GCC 5.4.0 20160609] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>>
>>>
>>> A = [(2, 'two', '2nd', 'second'), (3, 'three', '3rd', 'third')]
>>> B = ['three']
>>> [a[0] for b in B for a in A if b in a[1:]]
    [3]
>>> 
>>> 
>>> A = [(2, 'two', '2nd', 'second'), (3, 'three', '3rd', 'third')]
>>> B = ['two', '3rd']
>>> [a[0] for b in B for a in A if b in a[1:]]
    [2, 3]
>>> 

Answer 2

Try this single liner and use in to check element exitance: -

A = [(2, 'two', '2nd', 'second')]
B = ['two']
print [i[0] for i in A if B[0] in i]
#[2]
B = ['two1']
print [i[0] for i in A if B[0] in i]
#[]

Answer 3

You're almost there -- you just need

x for x,y,z,t in a

instead of

x for x,y,z,t in a[0]

Answer 4

Why not

A = [(2, 'two', '2nd', 'second')]
B = ['two']
if B[0] in A[0][1:]:
  print A[0][0]

Answer 5

Do not use list comprehension if you want to print. Use simple for loop.

In [12]: A = [(2, 'two', '2nd', 'second')]

In [13]: B = ['two']

In [14]: for tup in A:
    ...:     if B[0] in tup:
    ...:         print(tup[0])

If you want final list try list comprehension. You can index tuple. No need to unpack the entire tuple.

In [15]: [tup[0]
    ...:     for tup in A
    ...:         if B[0] in tup]
Out[15]: [2]

Answer 6

for x,y,z,t in a[0] is equal to for x,y,zt,t in 2,for x,y,zt in 'tow',for x,y,z,t in '2nd'……,but the 2 is int which is not iterable.You can try other ways such as cut the first element of a[0] than judge if a[0] is equal to b[0]. Hold I can help you!