我有1个主列表，它基于2个子列表。我想用“search_value”参数创建一个函数＆amp;想打印“search_value”项的索引位置，包括子列表的列表索引。
示例：
grocery = ["Juice", "Tomato", "Potato", "Banana", "Milk", "Bread"]
clothes = ["Shirt", "Pant", "Jacket", "Sweater", "Hat", "Pajama", "T-Shiraz", "Polo"]
master_list = [grocery, clothes]
预期结果：
"The Item You Searched For is", search_value, ". It is in the first/second list with index position of:", index #
我是python＆amp;的新手编写了工作代码。只想知道如何用更少的努力来做到这一点
def function(in_coming_string_to_search):
grocery = ["Juice", "Tomato", "Potato", "Banana", "Milk", "Bread"]
clothes = ["Shirt", "Pant", "Jacket", "Sweater", "Hat", "Pajama", "T-Shiraz", "Polo"]
master_list = [grocery, clothes]
length = int(len(master_list))
print master_list, "List has:", length, "list items", '\n'
to_do_list_first_array_index = 0
counter = 0
list_one_length = int(len(master_list[0]))
while counter < list_one_length:
for a in master_list[to_do_list_first_array_index]:
# print a
if a == in_coming_string_to_search:
print "The Item You Searched For is", in_coming_string_to_search, ". It is in the first list with index position of:", counter
counter = counter + 1
to_do_list_second_array_index = 1
counter2 = 0
list_two_length = int(len(master_list[1]))
while counter2 < list_two_length:
for b in master_list[to_do_list_second_array_index]:
if b == in_coming_string_to_search:
print "The Item You Searched For is", in_coming_string_to_search, ". It is in the second list with index position of:", counter2
counter2 = counter2 + 1
if __name__ == '__main__':
string_to_search = "Tomato"
function(string_to_search)
答案 0 :(得分：1)
假设master_list及其子列表在全局范围之前之前一次性定义
def search(needle):
for i, sublist in enumerate(master_list):
where = sublist.find(in_coming_string_to_search)
if where == -1: continue
print "The Item You Searched For is", needle
print "It is in the {} sublist, at {}".format(nd(i), where)
return
print "Could not find {} anywhere".format(needle)
ordinals = "first", "second", "third", "fourth", "fifth"
def nd(i):
if i<len(ordinals): return ordinals[i]
return "{}th".format(i)
答案 1 :(得分：1)
感谢所有帮助。我能够以更少的努力获得理想的结果。以下是我的最终代码。希望大家都同意：
def two_dim_list(incoming_item_to_search):
my_list = [["Banana", "Apple", "Orange", "Grape", "Pear"], ["Book", "Pen", "Ink", "Paper", "Pencil"], ["Shirt", "Pant", "Jacket", "Hat", "Coat"]]
list_length = len(my_list)
counter = 0
while counter < list_length:
try:
index = my_list[counter].index(incoming_item_to_search)
if index >= 0:
print "found item", incoming_item_to_search, "at index:", index, "of", counter, "sublist"
except ValueError:
pass
counter = counter + 1
if __name__ == '__main__':
item_to_search = "Coat"
two_dim_list(item_to_search)
答案 2 :(得分：0)
通过使用enumerate迭代子列表同时跟踪索引并使用每个子列表的.index方法，可以大大缩小这一点：
def FindInLists( listOfLists, searchTerm ):
for listIndex, subList in enumerate( listOfLists ):
if searchTerm in subList:
pos = subList.index( searchTerm )
print( "{term} found at position {pos} of list #{which}".format( term=repr( searchTerm ), pos=pos, which=listIndex ) )
return listIndex, pos
print( "{term} not found".format( term=repr( searchTerm ) ) )
return None, None
# test:
grocery = ["Juice", "Tomato", "Potato", "Banana", "Milk", "Bread"]
clothes = ["Shirt", "Pant", "Jacket", "Sweater", "Hat", "Pajama", "T-Shiraz", "Polo"]
master_list = [grocery, clothes]
print( FindInLists( master_list, "Hat" ) )
print( FindInLists( master_list, "Hats" ) )
答案 3 :(得分：0)
使用index功能，如上所述[{3}}：
try:
index1 = master_list[0].index(in_coming_string_to_search)
if index1 >= 0:
print "The Item You Searched For is", in_coming_string_to_search, ". It is in the first list with index position of:", index1
except ValueError:
pass
try:
index2 = master_list[1].index(in_coming_string_to_search)
if index2 >= 0:
print "The Item You Searched For is", in_coming_string_to_search, ". It is in the secpnd list with index position of:", index2
except ValueError:
pass
答案 4 :(得分：0)
以下功能将有所帮助：
def func(new_string):
grocery = ["Juice", "Tomato", "Potato", "Banana", "Milk", "Bread"]
clothes = ["Shirt", "Pant", "Jacket", "Sweater", "Hat", "Pajama", "T-Shiraz", "Polo"]
master_list = [grocery, clothes]
child_list_num = 1
counter = 0
for child_list in master_list:
for item in child_list:
if item == new_string:
print("The Item You Searched For is", new_string,
". It is in the chlid list "+ str(child_list_num) +
" with index position of:", counter)
return None
counter += 1
counter = 0
child_list_num += 1
print("The item does not exist.")
return None
if __name__ == '__main__':
item = "Sweater"
func(item)
如果您要打印该项目的所有实例，请删除＆＃39; return None＆＃39;在最里面的循环中的语句，添加一个found值，每当找到该项的第一个实例时，该值设置为1，并在结尾处添加以下语句：
if (found == 0):
print("The item does not exist.")
return None
答案 5 :(得分：0)
您的代码很难阅读并且非常冗余。
function作为功能名称，它是内置类型length = int(len(master_list))重复，使用len(master_list)，len将返回一个int，无需转换它。%）而不是使用逗号打印这里的代码相同。
def find_string(target):
master_list = [
["Juice", "Tomato", "Potato", "Banana", "Milk", "Bread"],
["Shirt", "Pant", "Jacket", "Sweater", "Hat", "Pajama", "T-Shiraz", "Polo"]
]
print "Master List has %d sub-lists.\n" % len(master_list)
print "The item you are searching for is '%s'." % target
for list_index, sub_list in enumerate(master_list):
for item_index, item in enumerate(sub_list):
if item == target:
print "\tFound at sub-list %d in index position %d" % (list_index+1, item_index)
break
if __name__ == '__main__':
find_string("Tomato")
输出：
Master List has 2 sub-lists.
The item you are searching for is 'Tomato'.
Found at sub-list 1 in index position 1
答案 6 :(得分：0)
您需要缩进代码。虽然通常不需要循环，特别是对于这样的情况（例如，您只是尝试遍历列表以匹配项目）。
查看我在这里使用的一些函数：例如，format将帮助清除在需要在字符串中表达多个变量时使用的语法。没有必要，但在某些情况下，这将是非常有益的。
在使用enumerate时，另请查看for loops以获取项目本身的索引。
最后，请查看PEP8，了解有关样式和格式的一些指导。简单和简短的变量名称总是最好的。只要它仍然清楚地表达了数据类型。
# In Python, you need to indent the contents of the function
# (the convention is 4 spaces)
def searchItem(searchString): # Avoid really long variable names.
grocery = [
"Juice", "Tomato", "Potato",
"Banana", "Milk", "Bread"
]
clothes = [
"Shirt", "Pant", "Jacket", "Sweater",
"Hat", "Pajama", "T-Shiraz", "Polo"
]
masterDict = {'grocery': grocery, 'clothes': clothes}
# No need to use int(). len() returns an int by default.
length = len(masterDict)
# Printing the master_list variable actually prints the list
# representation (and not the variable name), which is probably
# not what you wanted.
print 'master_dict has a length of {0}'.format(length), '\n'
itemInfo = None
for listType, subList in masterDict.iteritems():
for itemIndex, item in enumerate(subList):
if item == searchString:
itemInfo = {'listType': listType, 'itemIndex': itemIndex}
break
if itemInfo:
print ("The item you searched for is {0}. It's in the {1} "
"list with an index of {2}").format(
searchString, itemInfo['listType'], itemInfo['itemIndex'])
else:
print ('The item you searched for ("{0}") was not found.'
.format(searchString))
输入：
searchString = "Tomato"
searchItem(searchString)
<强>输出：
"The item you searched for is Tomato. It's in the grocery list with an index of 1"
输入：
searchString = "banana"
searchItem(searchString)
<强>输出：
'The item you searched for ("banana") was not found.'
答案 7 :(得分：0)
最简单的方法是：
grocery = ["Juice", "Tomato", "Potato", "Banana", "Milk", "Bread"]
clothes = ["Shirt", "Pant", "Jacket", "Sweater", "Hat", "Pajama", "T-Shiraz", "Polo"]
master_list = [grocery, clothes]
def search(needle):
return_string = "The item you searched for is {}. It is in the {}{} list with index position of {}"
ordinals = {"1":"st", "2":"nd", "3":"rd"}
for lst_idx,sublst in enumerate(master_list, start=1):
try:
needle_idx = str(sublst.index(needle))
except ValueError:
continue
else:
lst_idx = str(lst_idx)
return return_string.format(needle,
lst_idx,
ordinals.get(lst_idx, 'th'),
needle_idx)
return "The item {} is not in any list.".format(needle)
样本
In [11]: search("Juice")
Out[11]: 'The item you searched for is Juice. It is in the 1st list with index p
osition of 0'
In [12]: search("Pajama")
Out[12]: 'The item you searched for is Pajama. It is in the 2nd list with index
position of 5'
In [13]: search("Not existent")
Out[13]: 'The item Not existent is not in any list.'