python - 使用 else pass 的列表理解

Question

如何在列表理解中执行以下操作？

test = [["abc", 1],["bca",2]]

result = []
for x in test:
    if x[0] =='abc':
        result.append(x)
    else:
        pass
result
Out[125]: [['abc', 1]]

尝试 1:

[x if (x[0] == 'abc') else pass for x in test]
  File "<ipython-input-127-d0bbe1907880>", line 1
    [x if (x[0] == 'abc') else pass for x in test]
                                  ^
SyntaxError: invalid syntax

尝试 2:

[x if (x[0] == 'abc') else None for x in test]
Out[126]: [['abc', 1], None]

尝试 3:

[x if (x[0] == 'abc') for x in test]
  File "<ipython-input-122-a114a293661f>", line 1
    [x if (x[0] == 'abc') for x in test]
                            ^
SyntaxError: invalid syntax

Answer 1

if 需要放在最后，您不需要在列表理解中使用 pass。仅当满足 if 条件时才会添加该项目，否则将忽略该元素，因此 pass 在列表理解语法中隐式实现。

[x for x in test if x[0] == 'abc']

为了完整起见，此语句的输出为:

[['abc', 1]]