Question

我正在尝试使用下面的Python代码输出新列表。 print（单词）的输出应为['my','name','is','michael','apples','i','like','cars']。

目前，print (words)仅输出['cars']。我在这里缺少什么？

a = 'my name is michael and i like cars'
b = a.split()
words = None
for i, j in enumerate(b):
    words = []
    if j == "and" and b[i+1][0] == "i":
        words.append("apples")
    else:
        words.append(j)
print (words)

Answer 1

在循环外创建words，您只看到最后一个单词，因为每次迭代都会将单词设置为空列表：

words = [] # outside the loop
for i, j in enumerate(b):

如果and恰好是最后一个字，您还会得到IndexError。您可以在枚举中将起始索引设置为1，然后您不需要+1并且将避免任何潜在的错误索引：

words = []
for i, j in enumerate(b, 1):
    if j == "and" and b[i][0] == "i":

你可以把它全部放在列表理解中：

a = 'my name is michael and i like cars'
b = a.split()
words = ["apples" if wrd == "and" and b[i][0] == "i" else wrd for i, wrd in enumerate(b,1)]
print(words)
['my', 'name', 'is', 'michael', 'apples', 'like', 'cars']

您还可以使用iter和next来避免编制索引：

a = 'my name is michael and i like cars'
it = iter(a.split())
words = ["apples" if wrd == "and" and next(it," ")[0] == "i" else wrd for wrd in it ]
print(words)
['my', 'name', 'is', 'michael', 'apples', 'like', 'cars']

Answer 2

你的循环中存在问题。每次迭代时，都会将words重置为[]

将列表设置为for循环之外的[]，如下所示：

words = []
for i, j in enumerate(b):
    if j == "and" and b[i+1][0] == "i":
        words.append("apples")
    else:
        words.append(j)

这应该有用。

Answer 3

TigerhawkT3是正确的。要修复代码，不要将单词初始化为None，请尝试将其初始化为空数组：

a = 'my name is michael and i like cars'
b = a.split()
words = []
for i, j in enumerate(b):
    if j == "and" and b[i+1][0] == "i":
        words.append("apples")
    else:
        words.append(j)
print (words)

Answer 4

这应该适合你。

a = 'my name is michael and i like cars'
b = a.split()
words = []
for x in range(len(b)-1):
    if b[x] == "and" and b[x+1] == "i":
        words.append("apples")
    else:
        words.append(j)
print (words)

Answer 5

我有一个更紧凑的方法。

我认为它更易读且易于理解。您可以参考如下：

这是您的var I delcare回复：

a = 'my name is michael and i like cars'
b = a.split()
words = []
for i, j in enumerate(b):
    if j == "and" and b[i+1][0] == "i":
        words.append("apples")
    else:
        words.append(j)
print (words)

根据当前项目和下一项目的值替换列表中的值

5 个答案: