Python|根据第 N 个元组元素替换元组

有时，在处理数据时，我们可能会遇到需要替换与特定数据条目匹配的条目的问题。这可以是匹配的电话号码、ID 等。这在 Web 开发领域有它的应用。让我们讨论可以执行此任务的某些方式。

方法 #1：使用循环 + enumerate()
可以使用循环和枚举函数的组合来执行此任务，这有助于访问第 N 个元素，然后在满足条件时检查和替换。

# Python3 code to demonstrate working of
# Replace tuple according to Nth tuple element
# Using loops + enumerate()
  
# Initializing list
test_list = [('gfg', 1), ('was', 2), ('best', 3)]
  
# printing original list
print("The original list is : " + str(test_list))
  
# Initializing change record
repl_rec = ('is', 2)
  
# Initializing N 
N = 1
  
# Replace tuple according to Nth tuple element
# Using loops + enumerate()
for key, val in enumerate(test_list):
    if val[N] == repl_rec[N]:
        test_list[key] = repl_rec
        break
  
# printing result
print("The tuple after replacement is : " + str(test_list))

输出 ：

The original list is : [('gfg', 1), ('was', 2), ('best', 3)]
The tuple after replacement is : [('gfg', 1), ('is', 2), ('best', 3)]

方法#2：使用列表推导
这是解决此特定问题的单线方法。在此，我们只是迭代列表元素并保持匹配匹配的元组第 N 个元素并执行替换。

# Python3 code to demonstrate working of
# Replace tuple according to Nth tuple element
# Using list comprehension
  
# Initializing list
test_list = [('gfg', 1), ('was', 2), ('best', 3)]
  
# printing original list
print("The original list is : " + str(test_list))
  
# Initializing change record
repl_rec = ('is', 2)
  
# Initializing N 
N = 1
  
# Replace tuple according to Nth tuple element
# Using list comprehension
res = [repl_rec if sub[N] == repl_rec[N] else sub for sub in test_list]
  
# printing result
print("The tuple after replacement is : " + str(res))

输出 ：

The original list is : [('gfg', 1), ('was', 2), ('best', 3)]
The tuple after replacement is : [('gfg', 1), ('is', 2), ('best', 3)]