如何使用pandas截断句子的左右部分

Question

将句子转换成一个单词列表，然后找到根字符串的索引，应该是这样做的：

sentence = "lack of association between the promoter polymorphism of the mtnr1a gene and adolescent idiopathic scoliosis"
root = "mtnr1a"

try:
    words = sentence.split()
    n = words.index(root)
    cutoff = ' '.join(words[n-4:n+5])
except ValueError:
    cutoff = None

print(cutoff)

结果：

promoter polymorphism of the mtnr1a gene and adolescent idiopathic

如何在pandas数据帧中实现它？

我试着：

sentence = data['sentence'] 
root = data['rootword'] 
def cutOff(sentence,root): 
   try: 
      words = sentence.str.split() 
      n = words.index(root) 
      cutoff = ' '.join(words[n-4:n+5]) 
except ValueError: 
      cutoff = None 
      return cutoff 
data.apply(cutOff(sentence,root),axis=1)

但它不起作用。

编辑：

当词根在句子中的第一个位置时，当词根在句子中的最后一个位置时，如何在词根之后的4个字符串之后切分句子？例如：

sentence = "mtnr1a lack of association between the promoter polymorphism of the gene and adolescent idiopathic scoliosis"
out if root in first position:
"mtnr1a lack of association between"
out if root in last position:
"lack of association between the promoter polymorphism of the gene and adolescent idiopathic scoliosis"
"adolescent idiopathic scoliosis mtnr1a"

Answer 1

代码中的两个小调整应该可以解决您的问题：

首先，在数据帧上调用apply()会将该函数应用于调用该函数的dataframe的每一行中的值。

您不必将列作为输入传递给函数，而且调用sentence.str.split()也没有意义。在cutOff()函数中，sentence只是一个常规字符串(不是列)。

将函数更改为：

def cutOff(sentence,root): 
    try: 
        words = sentence.split()  # this is the line that was changed
        n = words.index(root) 
        cutoff = ' '.join(words[n-4:n+5]) 
    except ValueError: 
        cutoff = None 
    return cutoff

接下来，您只需指定将作为函数输入的列-您可以使用lambda完成此操作

df.apply(lambda x: cutOff(x["sentence"], x["rootword"]), axis=1)
#0    promoter polymorphism of the mtnr1a gene and a...
#dtype: object