[英]looping backward or forward in python list to find a match
我有一个 python 列表,我将在其中搜索并找到一个术语。 一旦我找到它,我需要在列表中向后移动并使用=
找到第一次出现,然后前进并使用;
找到第一次出现;
.
我尝试使用 while 循环,但它不起作用。
extract = [1,2,"3=","fd","dfdf","keyword","ssd","sdsd",";","dds"]
indices = [i for i,s in enumerate(extract) if 'keyword' in s]
for ind in indices:
ind_while_for = ind
ind_while_back = ind
if ('=' in extract[ind]) & (';' in extract[ind]):
print(extract[ind])
if (';' in extract[ind]) & ('=' not in extract[ind]):
while '=' in extract[ind_while_back-1]:
ind_while_back -= 1
print(' '.join(extract[ind_while_back:ind]))
结果要求: 3= fd dfdf keyword ssd sdsd ;
查找关键字的位置:
kw = extract.index("keyword")
在原始列表的子列表中,在关键字位置之前,找到包含"="
索引最大的元素:
eq = max(i for i,w in enumerate(extract[:kw])
if isinstance(w,str) and "=" in w)
找到包含";"
索引最小的元素在从前一个元素到末尾的子列表中:
semi = min(i for i,w in enumerate(extract[eq:], eq)
if isinstance(w,str) and ';' in w)
提取两个极端之间的子列表:
extract[eq:semi+1]
#['3=', 'fd', 'dfdf', 'keyword', 'ssd', 'sdsd', ';']
您可以使用:
l = [1, 2, "3=", "fd", "dfdf", "keyword", "ssd", "sdsd", ";", "dds"]
s = "keyword"
def take(last, iterable):
l = []
for x in iterable:
l.append(x)
if last in x:
break
return l
# get all elements on the right of s
right = take(';', l[l.index(s) + 1:])
# get all elements on the left of s using a reversed sublist
left = take('=', l[l.index(s)::-1])
# reverse the left list back and join it to the right list
subl = left[::-1] + right
print(subl)
['3=', 'fd', 'dfdf', 'keyword', 'ssd', 'sdsd', ';']
尝试以下功能:
extract = ['1','2','3=','fd','dfdf','keyword','ssd','sdsd',';','dds']
def get_output_list(extract, key):
equals_indices = [i for i,j in enumerate(extract) if '=' in j]
semicolon_indices = [i for i,j in enumerate(extract) if ';' in j]
if key not in extract or len(equals_indices) == 0 or len(semicolon_indices) == 0:
return 'no match found1'
keyword_index = extract.index(key)
if any([keyword_index<i for i in semicolon_indices]) and any([keyword_index>i for i in equals_indices]) :
required_equal_index = keyword_index - equals_indices[0]
required_semicolon_index = semicolon_indices[0] - keyword_index
for i in equals_indices:
if (i < keyword_index) and required_equal_index > i:
required_equal_index = i
for i in semicolon_indices:
if (i > keyword_index) and (required_semicolon_index < i) :
required_semicolon_index = i
return extract[required_equal_index:required_semicolon_index+1]
else :
return 'no match found'
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