Question

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

我得到：[] ['apple', 'Pie', 'is', 'Yummy']。

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

我得到：['Apple', 'Pie', 'is', 'Yummy'] []

它的行为就像我的条件运算符只在for循环的第一次迭代中读取一次，然后额外的迭代不会评估条件。

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;您想检查tidbit，而不是str：

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

我改为使用Python自己的str.isupper()测试来测试tidbit的第一个字符：

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

接下来，只需使用列表推导创建两个列表，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'])

帮助（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.

Python for循环迭代列表

3 个答案: