参数数量可变的 f 字符串？

Question

我需要创建类似的消息

Hi John, it's time to eat quantity_1 of food1, quantity_2 of food_2 ..., qunatity_n of food_n.

为此我得到了pandas数据帧，它每隔一段时间就会更新一次。例如，数据框有时看起来像

df1=

qantity,food
1,apple

有时喜欢

df2=

quantity,food
1,apple
3,salads
2,carrots

我需要在每次新更新时从数据帧中创建消息字符串。对于

df1

，f 字符串可以很好地工作，并且为了创建我想要的

Hi John, it's time to eat 1 apple.

消息，我可以这样做：

f"Hi John, it's time to eat {df.quantity} {df1.food}

如果我有类似

df2

的内容，并且我不明确知道

df2

有多长，我希望收到类似的消息

Hi John, it's time to eat 1 apple, 3 salads, 2 carrots.

如何创建这样的字符串？我想过使用“splat”运算符来实现类似

join(*zip(df.quantity, df.food))

之类的操作，但我还没有弄清楚。 tnx

Answer 1

试试这个：

result=','.join([str(i[0])+' '+i[1] for i in zip(df.quantity, df.food)])

print(result)

'1 apple, 2 salads, 3 carrots'

你可以添加这个以获得最终结果：

"Hi John, it's time to eat " + result

Hi John, it's time to eat 1 apple, 2 salads, 3 carrots

Answer 2

import pandas as pd 
df = pd.DataFrame([[1, 'apple'], [2, 'salads'], [5, 'carrots']], columns=['quantity','food'])
menu =  ', '.join(f"{quantity} {food}" for idx, (quantity, food) in df.iterrows())
print(f"Hi John, it's time to eat {menu}.")

输出

Hi John, it's time to eat 1 apple, 2 salads, 5 carrots

.

使用包inflect你可以用更好的语法来做到这一点：

import inflect

p=inflect.engine()
menu = p.join([f"{quantity} {food}" for idx, (quantity, food) in df.iterrows()])
print(f"Hi John, it's time to eat {menu}.")

输出：

Hi John, it's time to eat 1 apple, 2 salads, and 5 carrots.

变形甚至可以构造正确的单数/复数形式

Answer 3

有两种方法可以解决这个问题。 第一个选项是在数据框中创建消息列

df = pd.DataFrame(data={'quantity':  [1],'food': ['apple']})
df['message'] = df.apply(lambda x: f"Hi John, it's time to eat {x.quantity} {x.food}", axis = 1)
print(df['message'])

第二个选项是按索引对数据框对象进行切片，以在数据框之外创建消息

f"Hi John, it's time to eat {df.quantity[0]} {df.food[0]}"

要处理数据框中的多个记录，您可以迭代行

"Hi John, it's time to eat " + ", ".join(list((f"{df.quantity[i]} {df.food[i]}" for i in df.index)))

Answer 4

对于复杂的场景，我更喜欢使用 str.format()，因为它使代码更易于阅读。

在这种情况下：

import pandas as pd
df2=pd.DataFrame({'quantity':[1,3,2],'food':['apple','salads','carrots']})
def create_advice(df):
    s="Hi John, it's time to eat "+", ".join(['{} {}'.format(row['quantity'],row['food']) for index,row in df.iterrows()])+'.'
    return s

create_advice(df2)
>"Hi John, it's time to eat 1 apple, 3 salads, 2 carrots."

您可能还想在创建字符串之前稍微修改 df：

list_of_portioned_food=['salads']
df2['food']=df2.apply(lambda row: ('portions of ' if row['food'] in list_of_portioned_food else '')+row['food'],axis=1)
df2.iloc[len(df2)-1,1]='and '+str(df2.iloc[len(df2)-1,1])

再次应用上述功能：

create_advice(df2)
> "Hi John, it's time to eat 1 apple, 3 portions of salads, and 2 carrots."

Answer 5

试试这个

df1 = pd.DataFrame({'size':['1','2'], 'Food':['apple', 'banana']})
l_1 = [x + '' + y for x, y in zip(df1['size'], df1['Food'])]
"Hi John, it's time to eat " + ", ".join(l_1)

Answer 6

我实际上在我的项目中使用了 SQLite3，但解决方案仍然有效：您可以创建一个串联函数，循环遍历参数，将它们附加到字符串，并将生成的完整字符串插入到另一个 f 字符串中。

def concat_String(values):
    str = ""
    for v in values:
        str += f" {v}"

def create_New_Table(con, cur, table_name, values):
    try:
        cur.execute(f"CREATE TABLE {table_name}({concat_String(values)})")