Question

thelist = [{'color':'green', 'time':4}, {'color':'red','time':2},{'color':'blue','time':5}]

我怎么说：

If "red" is in thelist and time does not equal 2 for that element (that's we just got from the list):

Answer 1

使用any()查明是否存在满足条件的元素：

>>> any(item['color'] == 'red' and item['time'] != 2  for item in thelist)
False

Answer 2

def colorRedAndTimeNotEqualTo2(thelist):
    for i in thelist:
        if i["color"] == "red" and i["time"] != 2:
            return True
    return False

print colorRedAndTimeNotEqualTo2([{'color':'green', 'time':4}, {'color':'red','time':2},{'color':'blue','time':5}])

对于列表中的i遍历列表，将当前元素分配给i并执行块中的其余代码（对于i的每个值）

感谢Benson。

Answer 3

您可以在列表推导中执行大部分列表操作。这是一个为颜色为红色的所有元素列出时间的列表。然后你可以询问那些时候是否存在2。

thelist = [{'color':'green', 'time':4}, {'color':'red','time':2},{'color':'blue','time':5}]
reds = ( x['time'] == 2 for x in thelist if x['color'] == red )
if False in reds:
  do_stuff()

你可以通过消除这样的变量“reds”来进一步压缩它：

thelist = [{'color':'green', 'time':4}, {'color':'red','time':2},{'color':'blue','time':5}]
if False in ( x['time'] == 2 for x in thelist if x['color'] == red ):
  do_stuff()

Answer 4

嗯，没有什么比“查找”更优雅，但你可以使用列表理解：

matches = [x for x in thelist if x["color"] == "red" and x["time"] != 2]
if len(matches):
    m = matches[0]
    # do something with m

但是，我发现[0]和len（）很乏味。我经常使用带有数组切片的for循环，例如：

matches = [x for x in thelist if x["color"] == "red" and x["time"] != 2]
for m in matches[:1]:
    # do something with m

Answer 5

list = [{'color':'green', 'time':4}, {'color':'red','time':2},{'color':'blue','time':5}]
for i in list:
  if i['color'] == 'red' && i['time'] != 2:
    print i

Answer 6

for val in thelist:
    if val['color'] == 'red' and val['time'] != 2:
        #do something here

但它看起来不像是正确的数据结构。

如何确定元素是否在列表中？

6 个答案: