编写一个python程序，显示两个字符串之间的所有公共字符

Question

def firstAndLast(first,last):
    common = []
    if first in last:
        print first

first = list(raw_input("Enter first name: "))
last = list(raw_input("Enter last name: "))

firstAndLast(first, last)

ERROR:

File "<ipython-input-13-f4ec192dd3a8>", line 11
    print names
              ^
SyntaxError: Missing parentheses in call to 'print'. Did you mean print(names)?

Answer 1

The error tells you what to do. In python-3.x print() is function is with parentheses, so you've to pass the arguments in the parentheses.

print(first)

And you can also just use input() function instead of raw_input() function

PROGRAM

def firstAndLast(first,last):
    common = []
    for letter in first:
        if letter in last:
            common.append(letter)
    return common

first = input("Enter first name: ")
last = input("Enter last name: ")

common = firstAndLast(first, last)
#common will be the list of common letters

print(common)

OUTPUT

Enter first name: janee
Enter last name: denver
['n', 'e', 'e'] 

Answer 2

Program:

s1=raw_input("Enter first string:")

s2=raw_input("Enter second string:")

a=list(set(s1)&set(s2))

print("The common letters are:")

for i in a:
    print(i)

Answer 3

s1=input("Enter first string:")

s2=input("Enter second string:")

a=list(set(s1)&set(s2))

print("The common letters are:")

for i in a: print(i)

Answer 4

Try this,

first = list(input("Enter first name: "))
last = list(input("Enter last name: "))

print(f"common characters are: {', '.join(set(first) & set(last))}")

output:

Enter first name: Jaydip
Enter last name: Kyada
common characters are: a, y, d

I think this single line will answer you question.

Answer 5

You can numpy'sintersect1D :-

import numpy as np
def firstAndLast(first,last):
    return np.intersect1d(first, last)

first = input("Enter first name: ")
last = input("Enter last name: ")
first, last = [*first], [*last]

firstAndLast(first, last)

Output

Enter first name: gaurav
Enter last name: chowdhary
array(['a', 'r'], dtype='<U1')