問(wèn)題描述

File 'log1.py', line 140 for ref, occurnum in nsmallest(10, d.iteritems(), key=lambda (k,v): (-v,k)): ^SyntaxError: invalid syntax

代碼如下

def TopOccurr(num, request, total):args = parse_args()d = {}for i in get_obj[num]:if i in d:d[i] = d[i]+1else:d[i] = 1x = PrettyTable([request, total])x.align['Requests'] = 'l'x.padding_width = 1for ref, occurnum in nsmallest(10, d.iteritems(), key=lambda (k,v): (-v,k)):x.add_row([ref[:120], occurnum])return x.get_string(start=0, end=10, sortby=total, reversesort=True)

py3.5.2

問(wèn)題解答

lambda定義的時(shí)候，如果有多個(gè)參數(shù)，那么這些參數(shù)是不需要加括號(hào)的e.g.

l=lambda x,y:(y,x) # 正確

l=lambda (x,y):(y,x) # 報(bào)錯(cuò)

官方文檔中的lambda：

貼上修改后完整代碼：

def TopOccurr(num, request, total): args = parse_args() d = {} for i in get_obj[num]:if i in d: d[i] = d[i]+1else: d[i] = 1 x = PrettyTable([request, total]) x.align['Requests'] = 'l' x.padding_width = 1 for ref, occurnum in nsmallest(10, d.iteritems(), key=lambda k,v:(-v,k)):x.add_row([ref[:120], occurnum]) return x.get_string(start=0, end=10, sortby=total, reversesort=True)

實(shí)測(cè)沒(méi)有報(bào)錯(cuò)（Python 3.6.0，IPython 5.3.0）

python3中l(wèi)ambda不支持用括號(hào)的方式解壓數(shù)據(jù)

把參數(shù)的括號(hào)去掉就行了。