在python中为数据框中的特定单词赋值答案

【问题标题】：assigning a value to specific words in a dataframe in python在python中为数据框中的特定单词赋值
【发布时间】：2020-08-22 14:50:16
【问题描述】：

您好，我有一个由 7989 行 × 1 列组成的数据框。不同的行是不同海上海盗袭击的后果。

然后，我想根据特定单词是否包含在下面的不同列表之一中，为不同的行分配一个值。然后分配的值将取决于不同的列表。

6 个列表：

five =['kill','execute','dead']
four =['kidnap','hostag','taken','abduct']
three =['injur','wound','assault']
two =['captur','hijack']
one =['stolen','damage','threaten','robber','destroy']
zero =['alarm','no','none']

我尝试过这样做：

df['five']=df.apply(lambda x: '5' if x == 'five' else '-')

df 是我的数据框

谁能帮忙？

【问题讨论】：

你能在你的问题中嵌入一个简短的例子吗？

标签： python pandas list dataframe nlp

【解决方案1】：

您可以为每个列表创建带有数字值merge all dictionaries together 的字典，然后通过numpy.where 设置新列：

df = pd.DataFrame({'outcom':[['kill','dead'],['abduct','aaaa'],['hostag']]})

#same way add another lists
five = ['kill','execute','dead']
four = ['kidnap','hostag','taken','abduct']   
three =['injur','wound','assault']
two =['captur','hijack']
one =['stolen','damage','threaten','robber','destroy']
zero =['alarm','no','none']    

#same way add another dicts
d5 = dict.fromkeys(five, '5')
d4 = dict.fromkeys(four, '4')
d3 = dict.fromkeys(three, '3')
d2 = dict.fromkeys(two, '2')
d1 = dict.fromkeys(one, '1')
d0 = dict.fromkeys(zero, '0')

d = {**d5, **d4, **d3, **d2, **d1, **d0}
print (d)

for k, v in d.items():
    df[k] = np.where(df['outcom'].apply(lambda x: k in x), v, '-')

print (df)
           outcom kill execute dead kidnap hostag taken abduct
0    [kill, dead]    5       -    5      -      -     -      -
1  [abduct, aaaa]    -       -    -      -      -     -      4
2        [hostag]    -       -    -      -      4     -      -

【讨论】：

【解决方案2】：

编辑

您可以像这样使用loc 函数（documentation）：

将熊猫导入为 pd

five = ["I", "like"]
df = pd.DataFrame(["I", "like", "bacon", "in", "the", "morning"], columns=["Words"])
     Words
0        I
1    likes
2    bacon
3       in
4      the
5  morning

df["New"] = df["Words"].copy()
df.loc[df["New"] == "I", "New"] = 5

     Words      New
0        I        5
1     like     like
2    bacon    bacon
3       in       in
4      the      the
5  morning  morning

然后您可以使用 for 循环来帮助您

【讨论】：

非常感谢！但是我会创建一个新列，其中我的值是 0-5，具体取决于是否包含该单词。而且不能说 df['new_name_column'] = df.where(df ==['kidnap','hostag','taken','abduct], 4)
我想我也可以做一个循环，但它不起作用。 ://
那可能是因为我犯了一个错误，不像numpy pandas的@987654324@替换了DONT持有条件的值
@SofieAmalieBodenhoff 我修正了我的答案并运行了代码，它可以工作:)
我试过了，还是不行。我有一列有多个单词是这个问题吗？

【解决方案3】：

感谢大家的帮助，我想我找到了让它工作的方法：

 list_of_words = zero + one + two + three + four + five

     outcome_refined = df_Stop2['outcome'].apply(lambda x: [item for item in x if item 
      in list_of_words])



 outcome_numbered=[] #Create an empty list

def max_val(list): #Ensures that then we only get the largest possible value

maximum_value = 0

for i in list:
   
   if i > maximum_value:

        maximum_value = i

return [maximum_value]

        
 #Make sure that you loop through each of the lists
    
 for words in outcome_refined:
    tmp = [] #Create a temprorary empty list
    for word in words:
      if word in zero:
          word = 0
      elif word in one:
          word = 1
      elif word in two:
          word = 2
      elif word in three:
          word = 3
      elif word in four:
          word = 4
      elif word in five:
          word = 5   
      tmp.append(word)
    tmp = max_val(tmp)
    outcome_numbered.append(tmp)


df_Stop['outcome_numbered']=outcome_numbered.copy()   

df_Stop

Finally working

【讨论】：