6

检查 python 中是否存在列表或字典的最简单方法是什么?

我正在使用以下内容,但这不起作用:

if len(list) == 0:
    print "Im not here"

谢谢,

4

8 回答 8

10

您可以使用 try/except 块:

try:
    #work with list
except NameError:
    print "list isn't defined"
于 2012-07-19T07:58:10.510 回答
9

对于列表:

if a_list:
    print "I'm not here"

dicts 也是如此:

if a_dict:
    print "I'm not here"
于 2012-07-19T08:00:01.357 回答
6

当您尝试引用不存在的变量时,解释器会引发NameError. 但是,依赖代码中变量的存在是不安全的(您最好将其初始化为 None 或其他东西)。有时我用这个:

try:
    mylist
    print "I'm here"
except NameError:
    print "I'm not here"
于 2012-07-19T08:01:04.340 回答
4

如果你能够命名它 - 它显然“存在” - 我假设你的意思是检查它是否“非空”......最pythonic的方法是使用if varname:. 请注意,这不适用于生成器/可迭代对象以检查它们是否返回数据,因为结果将始终为True.

如果您只想使用某个索引/键,那么只需尝试使用它:

try:
    print someobj[5]
except (KeyError, IndexError) as e: # For dict, list|tuple
    print 'could not get it'
于 2012-07-19T07:56:18.863 回答
2

例子:

mylist=[1,2,3]
'mylist' in locals().keys()

或者使用这个:

mylist in locals().values()
于 2012-07-19T11:02:07.630 回答
1

简单的控制台测试:

>>> if len(b) == 0: print "Ups!"
... 
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'b' is not defined

>>> try:
...     len(b)
... except Exception as e:
...     print e
... 
name 'b' is not defined

示例展示了如何检查列表是否包含元素:

alist = [1,2,3]
if alist: print "I'm here!"

Output: I'm here!

除此以外:

alist = []
if not alist: print "Somebody here?"

Output: Somebody here?

如果您需要检查列表/元组的存在/不存在,这可能会有所帮助:

from types import ListType, TupleType
a_list = [1,2,3,4]
a_tuple = (1,2,3,4)

# for an existing/nonexisting list
# "a_list" in globals() check if "a_list" is defined (not undefined :p)

if "a_list" in globals() and type(a_list) is ListType: 
    print "I'm a list, therefore I am an existing list! :)"

# for an existing/nonexisting tuple
if "a_tuple" in globals() and type(a_tuple) is TupleType: 
    print "I'm a tuple, therefore I am an existing tuple! :)"

如果我们需要避免in globals()也许我们可以使用这个:

from types import ListType, TupleType
try:
    # for an existing/nonexisting list
    if type(ima_list) is ListType: 
        print "I'm a list, therefore I am an existing list! :)"

    # for an existing/nonexisting tuple
    if type(ima_tuple) is TupleType: 
        print "I'm a tuple, therefore I am an existing tuple! :)"
except Exception, e:
    print "%s" % e

Output:
    name 'ima_list' is not defined
    ---
    name 'ima_tuple' is not defined

参考书目: 8.15。types — 内置类型的名称 — Python v2.7.3 文档 https://docs.python.org/3/library/types.html

于 2012-08-28T16:58:40.387 回答
0

检查(1)变量存在和(2)检查它是列表

try:
    if type(myList) is list:
        print "myList is list"
    else:
        print "myList is not a list"
except:
    print "myList not exist"
于 2015-08-04T12:07:40.700 回答
-1
s = set([1, 2, 3, 4])
if 3 in s:
    print("The number 3 is in the list.")
else:
    print("The number 3 is NOT in the list.")

您可以在此处找到更多相关信息:https ://docs.quantifiedcode.com/python-anti-patterns/performance/using_key_in_list_to_check_if_key_is_contained_in_a_list.html

于 2018-11-06T18:02:54.010 回答