我有一个字符串列表(包含大量的字符串和列表)。我希望从列表中仅选取特定的元素来构建字典。
l = ['aaa 0','bbb 1','ccc 2','ddd 3','abc 1']
wanted = set(['aaa','abc'])
dict([x.split() for x in l if x.split()[0] in wanted])
{'aaa': '0', 'abc': '1'}
在不调用split()
操作两次的情况下,这是否可以完成?
我有一个字符串列表(包含大量的字符串和列表)。我希望从列表中仅选取特定的元素来构建字典。
l = ['aaa 0','bbb 1','ccc 2','ddd 3','abc 1']
wanted = set(['aaa','abc'])
dict([x.split() for x in l if x.split()[0] in wanted])
{'aaa': '0', 'abc': '1'}
在不调用split()
操作两次的情况下,这是否可以完成?
使用嵌套的生成器表达式 ((x.split() for x in l)
) 来产生键值对:
>>> l = ['aaa 0', 'bbb 1', 'ccc 2', 'ddd 3', 'abc 1']
>>> wanted = {'aaa', 'abc'}
>>> {key: value for key, value in (x.split() for x in l) if key in wanted}
{'abc': '1', 'aaa': '0'}
split
:{k: n for k, n in dict([x.split() for x in l]).iteritems() if k in wanted}
l = ['aaa 0','bdb 1','ccc 2','ddd 3','abc 1']
dict = { }
s = set(['aaa','abc'])
for elem in l:
nl = elem.split()
if nl[0] in s:
dict[nl[0]] = nl[1]
for elm in dict:
print elm,dict[elm]