在Python 2.7.6中拾取列表中的元素

在Python2.7.6中，一个列表如下所示。以何种方式提取以“4”开头、长度为4的项目，即下面的4646和4648

aaa = [2013, 2014, 2002, 4646, 4648, 20, 456, 5623, 'abc']

我只能通过以下方式计算出选择4个长度：

results = []

for number in aaa:
  if len(str(number)) == 4:
      results.append(number)

print results

谢谢

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这些都是很好的答案。但我是新手，所以选择最简单的一个。：）

我想知道这是你想要的

results = []

for number in aaa:
  if str(number)[0] == '4' and len(str(number)) == 4:
      results.append(number)

print results

使用过滤器：

filtered = list(filter(lambda x: len(str(x)) == 4 and str(x)[0] == '4', aaa))

使用列表理解：

filtered = [ x for x in aaa if len(str(x)) == 4 and str(x)[0] == '4' ]

使用生成器表达式：

ge = ( x for x in aaa if len(str(x)) == 4 and str(x)[0] == '4' )
filtered = list(ge)

现在已过滤为：

[464648]

带整数：

result = [x for x in aaa if 4000 <= x < 5000]

只是：

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希望这能给你带来想要的结果

aaa = [2013, 2014, 2002, 4646, 4648, 20, 456, 5623, 'abc']
results = []
for number in aaa:
    if len(str(number)) == 4 and str(number).split()[0][0] == '4':
        results.append(number)
print results 

Output: [4646, 4648]

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请尝试以下代码行：

aaa = [2013, 2014, 2002, 4646, 4648, 20, 456, 5623, 'abc']

results = []

for number in aaa:
  if len(str(number)) == 4 and str(number)[0] == '4':
  results.append(number)

print results

我仅更改了您的if声明，如：

如果len（str（number））==4和str（number）[0]==4'：

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除非您可能也有以4开头的字符串，否则为什么不在

范围（40005000）

上进行筛选呢？

aaa = [2013, 2014, 2002, 4646, 4648, 20, 456, 5623, 'abc']
results = []
for number in aaa:
    if len(str(number)) == 4 and str(number).split()[0][0] == '4':
        results.append(number)
print results 

Output: [4646, 4648]

aaa = [2013, 2014, 2002, 4646, 4648, 20, 456, 5623, 'abc']

results = []

for number in aaa:
  if len(str(number)) == 4 and str(number)[0] == '4':
  results.append(number)

print results