[英]Python using “IN” in a “IF ELSE” loop
i have a list of tuples that i loop through in a simple for loop to identify tuples that contain some conditions. 我有一个元组列表,我在一个简单的for循环中循环遍历,以识别包含某些条件的元组。
mytuplist =
[(1, 'ABC', 'Today is a great day'), (2, 'ABC', 'The sky is blue'),
(3, 'DEF', 'The sea is green'), (4, 'ABC', 'There are clouds in the sky')]
I want it to be efficient and readable like this: 我希望它像这样高效且易读:
for tup in mytuplist:
if tup[1] =='ABC' and tup[2] in ('Today is','The sky'):
print tup
The above code does not work and nothing gets printed. 上面的代码不起作用,什么也没打印。
This code below works but is very wordy. 下面的代码可以工作,但是非常麻烦。 How do i make it like the above?
我如何像上面那样?
for tup in mytuplist:
if tup[1] =='ABC' and 'Today is' in tup[2] or 'The sky' in tup[2]:
print tup
You should use the built-in any()
function: 您应该使用内置的
any()
函数:
mytuplist = [
(1, 'ABC', 'Today is a great day'),
(2, 'ABC', 'The sky is blue'),
(3, 'DEF', 'The sea is green'),
(4, 'ABC', 'There are clouds in the sky')
]
keywords = ['Today is', 'The sky']
for item in mytuplist:
if item[1] == 'ABC' and any(keyword in item[2] for keyword in keywords):
print(item)
Prints: 印刷品:
(1, 'ABC', 'Today is a great day')
(2, 'ABC', 'The sky is blue')
您不需要,因为in
与序列仅匹配整个元素。
if tup[1] =='ABC' and any(el in tup[2] for el in ('Today is', 'The sky')):
Your second approach (which, however, needs to be parenthesized as x and (y or z)
to be correct) is necessary tup[2]
contains one of your key phrases, rather than being an element of your set of key phrases. 您的第二种方法(不过,必须将其用
x and (y or z)
括起来才能正确)括起来) tup[2]
包含您的一个关键短语,而不是您的一组关键短语中的一个。 You could use regular-expression matching at the cost of some performance: 您可以使用正则表达式匹配,但会降低性能:
if tup[1] == 'ABC' and re.match('Today is|The sky', tup[2])
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