[英]Python: for each tuple in list, check if string in tuple
I know how to cycle through all tuples in a list. 我知道如何遍历列表中的所有元组。 However, my problem is to see if a string is in a tuple within the list. 但是,我的问题是查看字符串是否在列表中的元组中。 I have constructed the following: 我构造了以下内容:
# where:
unformatted_returns = [('2015-6-10', u'88.48'), ('2015-6-9', u'86.73'), ('2015-6-8', u'86.15'), ('2015-6-5', u'86.05')]
date_new = '2015-6-8'
for n in unformatted_returns: # The problem is here
if str(date_new) in n[0]:
print "found date"
rate_of_return_calc(date, date_new)
pass
else:
print "date {0} not found".format(date_new)
day_accumulater(d, m, y, date_new, date)
The problem is that the first tuple in the cycle n in unformatted_returns
does not satisfy the condition, therefore it prints "not found". 问题在于unnated_returns中循环n in unformatted_returns
的第一个元组不满足条件,因此它打印“找不到”。 I don't want it do this obviously, because date_new
is actually in the list! 我不希望它这样做,因为date_new
实际上在列表中!
So, how do I get the program to cycle through each n
, then if all n
's do not satisfy containing date_new
then print "day not found"
? 因此,如何使程序循环遍历每个n
,然后如果所有n
不满足包含date_new
则打印"day not found"
?
Move the else
one level down . 将else
向下一级 。 for
loops also take an else
suite, which is executed when you did not exit the loop early . for
循环还采用了else
套件,该套件在您没有提前退出循环时执行。 Then add a break
: 然后添加一个break
:
for n in unformatted_returns:
if date_new == n[0]:
print "found date"
rate_of_return_calc(date, date_new)
break
else:
print "date {0} not found".format(date_new)
day_accumulater(d, m, y, date_new, date)
I've also cleaned up your test; 我也清理了你的考试; you are matching n[0]
against a date string, you want them to be equal , not for one to be a substring of the other. 您要将n[0]
与日期字符串进行匹配,希望它们相等 ,而不是让其中一个成为另一个的子字符串。
Now one of two things can happen: 现在可以发生以下两种情况之一:
date_new
is equal to the first element of one of the tuples. date_new
等于其中一个元组的第一个元素。 The break
is executed, the for
loop ends, and the else
suite is skipped. 执行break
, for
循环结束, else
套件被跳过。
date_new
is not equal to any of the first elements of the tuples. date_new
不等于元组的任何第一个元素。 The break
is never executed, the loop ends and the else
suite is executed to show no match was found. 决不执行break
,循环结束,执行else
套件以显示未找到匹配项。
Demo: 演示:
>>> unformatted_returns = [('2015-6-10', u'88.48'), ('2015-6-9', u'86.73'), ('2015-6-8', u'86.15'), ('2015-6-5', u'86.05')]
>>> date_new = '2015-6-8'
>>> for n in unformatted_returns:
... if date_new == n[0]:
... print "found date"
... break
... else:
... print "date {0} not found".format(date_new)
...
found date
>>> date_new = '2015-6-7' # not in the list
>>> for n in unformatted_returns:
... if date_new == n[0]:
... print "found date"
... break
... else:
... print "date {0} not found".format(date_new)
...
date 2015-6-7 not found
This obviously would only ever find the first such matching element. 显然,这只会找到第一个这样的匹配元素。
If you must process all matching elements, a flag is usually easiest: 如果必须处理所有匹配的元素,则通常最简单的标志是:
found = False
for n in unformatted_returns:
if date_new == n[0]:
print "found date"
rate_of_return_calc(date, date_new)
found = True
if not found:
print "date {0} not found".format(date_new)
day_accumulater(d, m, y, date_new, date)
All this assumes that n[1]
is of interest too. 所有这些都假定n[1]
也很重要。 If all you need to know if the date is present , use any()
and a generator expression to test for a matching element: 如果你需要知道,如果日期是目前使用any()
和发电机表达式来测试一个匹配的元素:
if any(n[0] == date_new for n in unformatted_returns):
print "found date"
rate_of_return_calc(date, date_new)
else:
print "date {0} not found".format(date_new)
day_accumulater(d, m, y, date_new, date)
Now we won't know which n
matched, but that doesn't actually matter. 现在我们不知道哪个 n
匹配,但这实际上并不重要。
Just use a flag: 只需使用一个标志:
unformatted_returns = [('2015-6-10', u'88.48'), ('2015-6-9', u'86.73'), ('2015-6-8', u'86.15'), ('2015-6-5', u'86.05')]
date_new = '2015-6-8'
found_one = False
for n in unformatted_returns: # The problem is here
if str(date_new) in n[0]:
print "found date"
rate_of_return_calc(date, date_new)
found_one = True
# add a break here if you are only interested in the first match
if not found_one:
print "date {0} not found".format(date_new)
day_accumulater(d, m, y, date_new, date)
Just search for matches and if the search comes with no results, nothing was found: 只是搜索匹配项,如果搜索没有结果,则找不到任何内容:
matching_returns = [ur for ur in unformatted_returns if date_new == ur[0]]
if not matching_returns:
print "date {0} not found".format(date_new)
else:
# do something with matching_returns
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