Python for循环遍历列表

Question

如果我输入"apple Pie is Yummy"，我想要：['Pie','Yummy'] ['apple','is']

我明白了：[] ['apple', 'Pie', 'is', 'Yummy']。

如果我输入"Apple Pie is Yummy"，我想要：['Apple','Pie','Yummy'] ['is']

我明白了：['Apple', 'Pie', 'is', 'Yummy'] []

它的行为就像我的条件运算符在for循环的第一次迭代期间只被读取一次，然后额外的迭代不会计算条件。

str = input("Please enter a sentence: ")

chunks = str.split()

# create tuple for use with startswith string method
AtoZ = ('A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z')

# create empty lists to hold data
list1 = []
list2 = []

for tidbit in chunks:
    list1.append(tidbit) if (str.startswith(AtoZ)) else list2.append(tidbit)

print(list1)
print(list2)

Answer 1

您正在测试错误的变量；您希望检查tidbit，而不是str

list1.append(tidbit) if (tidbit.startswith(AtoZ)) else list2.append(tidbit)

我转而使用Python自己的str.isupper()测试，只测试tidbit的第一个字符

list1.append(tidbit) if tidbit[0].isupper() else list2.append(tidbit)

接下来，只需使用列表理解创建这两个列表，因为使用条件表达式的副作用非常可怕：

list1 = [tidbit for tidbit in chunks if tidbit[0].isupper()]
list2 = [tidbit for tidbit in chunks if not tidbit[0].isupper()]

Answer 2

chunks = raw_input("Enter a sentence: ").split()
list1 = [chunk for chunk in chunks if chunk[0].isupper()]
list2 = [chunk for chunk in chunks if chunk not in list1]

Answer 3

您可以在此处使用str.isupper()：

def solve(strs):

    dic={"cap":[],"small":[]}

    for x in strs.split():
        if x[0].isupper():
            dic["cap"].append(x)
        else:    
            dic["small"].append(x)

    return dic["small"],dic["cap"]



In [5]: solve("apple Pie is Yummy")
Out[5]: (['apple', 'is'], ['Pie', 'Yummy'])

In [6]: solve("Apple Pie is Yummy")
Out[6]: (['is'], ['Apple', 'Pie', 'Yummy'])

help(str.upper)

In [7]: str.isupper?
Type:       method_descriptor
String Form:<method 'isupper' of 'str' objects>
Namespace:  Python builtin
Docstring:
S.isupper() -> bool

Return True if all cased characters in S are uppercase and there is
at least one cased character in S, False otherwise.