如果/否则列表理解

问题:如果/否则列表理解

如何在Python中执行以下操作?

row = [unicode(x.strip()) for x in row if x is not None else '']

实质上:

  1. 用空字符串替换所有的None,然后
  2. 执行功能。

How can I do the following in Python?

row = [unicode(x.strip()) for x in row if x is not None else '']

Essentially:

  1. replace all the Nones with empty strings, and then
  2. carry out a function.

回答 0

您完全可以做到这一点。这只是一个订购问题:

[unicode(x.strip()) if x is not None else '' for x in row]

一般来说,

[f(x) if condition else g(x) for x in sequence]

而且,if仅对于具有条件的列表理解而言,

[f(x) for x in sequence if condition]

请注意,这实际上使用了一种不同的语言构造,即条件表达式,它本身不是理解语法的一部分,而ifafter则for…in是列表理解的一部分,用于从可迭代的源中筛选元素。


条件表达式可用于要根据条件在两个表达式值之间进行选择的所有情况。这与其他语言中存在三元运算符?:相同。例如:

value = 123
print(value, 'is', 'even' if value % 2 == 0 else 'odd')

You can totally do that. It’s just an ordering issue:

[unicode(x.strip()) if x is not None else '' for x in row]

In general,

[f(x) if condition else g(x) for x in sequence]

And, for list comprehensions with if conditions only,

[f(x) for x in sequence if condition]

Note that this actually uses a different language construct, a conditional expression, which itself is not part of the comprehension syntax, while the if after the for…in is part of list comprehensions and used to filter elements from the source iterable.


Conditional expressions can be used in all kinds of situations where you want to choose between two expression values based on some condition. This does the same as the ternary operator ?: that exists in other languages. For example:

value = 123
print(value, 'is', 'even' if value % 2 == 0 else 'odd')

回答 1

单程:

def change(f):
    if f is None:
        return unicode(f.strip())
    else:
        return ''

row = [change(x) for x in row]

虽然您有:

row = map(change, row)

或者您可以使用lambda内联。

One way:

def change(f):
    if f is None:
        return unicode(f.strip())
    else:
        return ''

row = [change(x) for x in row]

Although then you have:

row = map(change, row)

Or you can use a lambda inline.


回答 2

这是另一个说明性示例:

>>> print(", ".join(["ha" if i else "Ha" for i in range(3)]) + "!")
Ha, ha, ha!

它利用了这样的事实if i计算结果为False0True所有其它值由函数生成的range()。因此,列表理解评估如下:

>>> ["ha" if i else "Ha" for i in range(3)]
['Ha', 'ha', 'ha']

Here is another illustrative example:

>>> print(", ".join(["ha" if i else "Ha" for i in range(3)]) + "!")
Ha, ha, ha!

It exploits the fact that if i evaluates to False for 0 and to True for all other values generated by the function range(). Therefore the list comprehension evaluates as follows:

>>> ["ha" if i else "Ha" for i in range(3)]
['Ha', 'ha', 'ha']

回答 3

前面的答案已经解决了特定的问题,因此我将解决在列表推导中使用条件语句的一般想法。

这是一个示例,显示了如何在列表推导中编写条件语句:

X = [1.5, 2.3, 4.4, 5.4, 'n', 1.5, 5.1, 'a']     # Original list

# Extract non-strings from X to new list
X_non_str = [el for el in X if not isinstance(el, str)]  # When using only 'if', put 'for' in the beginning

# Change all strings in X to 'b', preserve everything else as is
X_str_changed = ['b' if isinstance(el, str) else el for el in X]  # When using 'if' and 'else', put 'for' in the end

请注意,在的第一个列表理解中X_non_str,顺序为:

表达 项目 迭代 如果 条件

在的最后一个列表理解中X_str_changed,顺序为:

表达式 如果 条件 其他 表达式2 项目 迭代

我总是觉得很难记住expresseion1必须是之前,如果表达式2必须是经过别人。我的头希望两者都在之前或之后。

我想它的设计一样,因为它类似于正常的语言,例如:“我想留在里面,如果下雨,否则我要到外面去”

用普通的英语,上面提到的两种列表理解可以表述为:

仅使用if

extract_apple 苹果 box_of_apples 如果 apple_is_ripe

if/else

mark_apple, 如果 apple_is_ripe 其他 ,则box_of_apples 中将 苹果标记为未 标记

The specific problem has already been solved in previous answers, so I will address the general idea of using conditionals inside list comprehensions.

Here is an example that shows how conditionals can be written inside a list comprehension:

X = [1.5, 2.3, 4.4, 5.4, 'n', 1.5, 5.1, 'a']     # Original list

# Extract non-strings from X to new list
X_non_str = [el for el in X if not isinstance(el, str)]  # When using only 'if', put 'for' in the beginning

# Change all strings in X to 'b', preserve everything else as is
X_str_changed = ['b' if isinstance(el, str) else el for el in X]  # When using 'if' and 'else', put 'for' in the end

Note that in the first list comprehension for X_non_str, the order is:

expression for item in iterable if condition

and in the last list comprehension for X_str_changed, the order is:

expression1 if condition else expression2 for item in iterable

I always find it hard to remember that expresseion1 has to be before if and expression2 has to be after else. My head wants both to be either before or after.

I guess it is designed like that because it resembles normal language, e.g. “I want to stay inside if it rains, else I want to go outside”

In plain English the two types of list comprehensions mentioned above could be stated as:

With only if:

extract_apple for apple in box_of_apples if apple_is_ripe

and with if/else

mark_apple if apple_is_ripe else leave_it_unmarked for apple in box_of_apples


回答 4

其他解决方案非常适合单个if/ else结构。但是,列表理解内的三元语句很难理解。

使用功能有助于提高可读性,但是很难在以映射为输入的工作流中扩展或适应这种解决方案。字典可以缓解这些问题:

row = [None, 'This', 'is', 'a', 'filler', 'test', 'string', None]

d = {None: '', 'filler': 'manipulated'}

res = [d.get(x, x) for x in row]

print(res)

['', 'This', 'is', 'a', 'manipulated', 'test', 'string', '']

The other solutions are great for a single if / else construct. However, ternary statements within list comprehensions are arguably difficult to read.

Using a function aids readability, but such a solution is difficult to extend or adapt in a workflow where the mapping is an input. A dictionary can alleviate these concerns:

row = [None, 'This', 'is', 'a', 'filler', 'test', 'string', None]

d = {None: '', 'filler': 'manipulated'}

res = [d.get(x, x) for x in row]

print(res)

['', 'This', 'is', 'a', 'manipulated', 'test', 'string', '']

回答 5

它与列表理解的执行方式有关。

请记住以下几点:

[ expression for item in list if conditional ]

等效于:

for item in list:
    if conditional:
        expression

其中的expression格式略有不同(请考虑在句子中切换主语和动词顺序)。

因此,您的代码[x+1 for x in l if x >= 45]执行以下操作:

for x in l:
    if x >= 45:
        x+1

但是,此代码可以[x+1 if x >= 45 else x+5 for x in l]做到这一点(重新排列之后expression):

for x in l:
    if x>=45: x+1
    else: x+5

It has to do with how the list comprehension is performed.

Keep in mind the following:

[ expression for item in list if conditional ]

Is equivalent to:

for item in list:
    if conditional:
        expression

Where the expression is in a slightly different format (think switching the subject and verb order in a sentence).

Therefore, your code [x+1 for x in l if x >= 45] does this:

for x in l:
    if x >= 45:
        x+1

However, this code [x+1 if x >= 45 else x+5 for x in l] does this (after rearranging the expression):

for x in l:
    if x>=45: x+1
    else: x+5

回答 6

不需要三元if / then / else。我认为您的问题需要这个答案:

row = [unicode((x or '').strip()) for x in row]

There isn’t any need for ternary if/then/else. In my opinion your question calls for this answer:

row = [unicode((x or '').strip()) for x in row]

回答 7

从迭代列表中列出项目

最好首先概括所有可能的形式,而不是给出问题的具体答案。否则,读者将不知道答案是如何确定的。这是我在尝试确定是否可以在最后一种形式中使用final’子句之前想出的几种通用形式。

[expression1(item)                                        for item in iterable]

[expression1(item) if conditional1                        for item in iterable]

[expression1(item) if conditional1 else expression2(item) for item in iterable]

[expression1(item) if conditional1 else expression2(item) for item in iterable if conditional2]

的值item不需要在任何条件子句中使用。A conditional3可用作将值添加或不添加到输出列表的开关。

例如,要创建一个新列表以从原始字符串列表中消除空字符串或空格字符串:

newlist = [s for s in firstlist if s.strip()]

Make a list from items in an iterable

It seems best to first generalize all the possible forms rather than giving specific answers to questions. Otherwise, the reader won’t know how the answer was determined. Here are a few generalized forms I thought up before I got a headache trying to decide if a final else’ clause could be used in the last form.

[expression1(item)                                        for item in iterable]

[expression1(item) if conditional1                        for item in iterable]

[expression1(item) if conditional1 else expression2(item) for item in iterable]

[expression1(item) if conditional1 else expression2(item) for item in iterable if conditional2]

The value of item doesn’t need to be used in any of the conditional clauses. A conditional3 can be used as a switch to either add or not add a value to the output list.

For example, to create a new list that eliminates empty strings or whitespace strings from the original list of strings:

newlist = [s for s in firstlist if s.strip()]

回答 8

您可以在理解中结合条件逻辑:

 ps = PorterStemmer()
 stop_words_english = stopwords.words('english')
 best = sorted(word_scores.items(), key=lambda x: x[1], reverse=True)[:10000]
 bestwords = set([w for w, s in best])


 def best_word_feats(words):
   return dict([(word, True) for word in words if word in bestwords])

 # with stemmer
 def best_word_feats_stem(words):
   return dict([(ps.stem(word), True) for word in words if word in bestwords])

 # with stemmer and not stopwords
 def best_word_feats_stem_stop(words):
   return dict([(ps.stem(word), True) for word in words if word in bestwords and word not in stop_words_english])

You can combine conditional logic in a comprehension:

 ps = PorterStemmer()
 stop_words_english = stopwords.words('english')
 best = sorted(word_scores.items(), key=lambda x: x[1], reverse=True)[:10000]
 bestwords = set([w for w, s in best])


 def best_word_feats(words):
   return dict([(word, True) for word in words if word in bestwords])

 # with stemmer
 def best_word_feats_stem(words):
   return dict([(ps.stem(word), True) for word in words if word in bestwords])

 # with stemmer and not stopwords
 def best_word_feats_stem_stop(words):
   return dict([(ps.stem(word), True) for word in words if word in bestwords and word not in stop_words_english])

回答 9

# coding=utf-8

def my_function_get_list():
    my_list = [0, 1, 2, 3, 4, 5]

    # You may use map() to convert each item in the list to a string, 
    # and then join them to print my_list

    print("Affichage de my_list [{0}]".format(', '.join(map(str, my_list))))

    return my_list


my_result_list = [
   (
       number_in_my_list + 4,  # Condition is False : append number_in_my_list + 4 in my_result_list
       number_in_my_list * 2  # Condition is True : append number_in_my_list * 2 in my_result_list
   )

   [number_in_my_list % 2 == 0]  # [Condition] If the number in my list is even

   for number_in_my_list in my_function_get_list()  # For each number in my list
]

print("Affichage de my_result_list [{0}]".format(', '.join(map(str, my_result_list))))

(venv)$ python list_comp.py
my_list的
关联[0,1,2,3,4,5 ] my_result_list的关联[0,5,4,7,8,9]

因此,为您: row = [('', unicode(x.strip()))[x is not None] for x in row]

# coding=utf-8

def my_function_get_list():
    my_list = [0, 1, 2, 3, 4, 5]

    # You may use map() to convert each item in the list to a string, 
    # and then join them to print my_list

    print("Affichage de my_list [{0}]".format(', '.join(map(str, my_list))))

    return my_list


my_result_list = [
   (
       number_in_my_list + 4,  # Condition is False : append number_in_my_list + 4 in my_result_list
       number_in_my_list * 2  # Condition is True : append number_in_my_list * 2 in my_result_list
   )

   [number_in_my_list % 2 == 0]  # [Condition] If the number in my list is even

   for number_in_my_list in my_function_get_list()  # For each number in my list
]

print("Affichage de my_result_list [{0}]".format(', '.join(map(str, my_result_list))))

(venv) $ python list_comp.py
Affichage de my_list [0, 1, 2, 3, 4, 5]
Affichage de my_result_list [0, 5, 4, 7, 8, 9]

So, for you: row = [('', unicode(x.strip()))[x is not None] for x in row]