如何在 DataFrame 中将单词与数字相乘？

Question

我有一个这样的数据框：print(df.words[0])[('replacement', 1), ('shaver', 2)]print(df.words[1])[('filter', 2), ('purifier', 1), ('please', 2)]我想创建一个名为“all_words”的新列。该列应该代表真实的字符串，而不是数字。('head', 3) should be: "head,head,head"示例的所需输出：print(df.all_words[0])'replacement, shaver, shaver'print(df.all_words[1])'filter, filter, purifier, please, please'

Answer 1

您将需要apply一个函数来将元组连接到单个字符串。

df['all_words'] = df.words.apply(lambda x: ', '.join(', '.join([y[0]] * y[1]) for y in x))

Answer 2

你可以这样做df.apply()

将熊猫导入为 pd

df = pd.DataFrame({'words' : [[('replacement', 1), ('shaver', 2)], [('filter', 2), ('purifier', 1), ('please', 2)]]})

def word_to_words(row):

    words_string = ''

    for tuple_set in row['words']:

        words_string += (tuple_set[0] + ', ') * tuple_set[1]

    return(words_string)

df['all_words'] = df.apply(word_to_words, axis=1)

Answer 3

您可以使用apply：

df = pd.DataFrame(data=[[[('filter', 2), ('purifier', 1), ('please', 2)]]], columns=['words'])

result = df.words.apply(lambda x: ', '.join(word for word, count in x for _ in range(count)))

print(result)

输出

0    filter, filter, purifier, please, please

Name: words, dtype: object

Answer 4

它只是解决 thorug 循环及其工作原理。如果单词是多个元组的列表

words = [[('replacement', 2), ('shaver', 2) ], [('filters', 2), ('purifier',1), ('plears', 3) ]]

Loop = words[0] #here you use indexing of words[0] or [1] both

Result = ()

for val in Loop:

    v = tuple([val[0] * 1 for _ in range(val[1])])

    Result = Result + v

Print(Result)