计算每一行数据框中的关键字

Question

我想计算我的列表中每个关键字在每行的给定数据框列中出现的数量总和。

d = {
    'Column_1': ['mango pret Orange No manner', ' préts No scan '], 
    'Column_2': ['read priority No', 'This is a priority noir '],
    'Column_3': ['No add', 'yep']
}

df = pd.DataFrame(data=d)

list_1 = ['Apple', 'Mango' ,'Orange', 'pr[éeêè]t[s]?']
list_2 = ['weather', 'r[ea]d' ,'p[wr]iority', 'noir?']
list_3 = ['n[eéè]d','snow[s]?', 'blanc?']

dict = {
    "s1": ['Column_1', list_1],
    "s2": ['Column_1', list_3],
    "s3": ['Column_2', list_2],
    "s4": ['Column_3', list_3],
    "s5": ['Column_2','Column_3',list_1]
}

for elt in list(dict.keys()):
    #s1 s2 s3 print(elt)
    if len(dict[elt])<=2:
        d = Counter(re.findall(r'|'.join(dict[elt][1]).lower(), str(df[dict[elt][0]].str.lower())))
        print(d)
        #df[elt] = d 
        sum(d.values())
    elif len(dict[elt])>2:
        aa = Counter(re.findall(r'|'.join(dict[elt][2]).lower(), str(df[dict[elt][0]].str.lower())))
        bb = Counter(re.findall(r'|'.join(dict[elt][2]).lower(), str(df[dict[elt][1]].str.lower())))
        b = sum(bb.values()) 
        a = sum(aa.values()) 
        d = a +b 
        df[elt] = d

我print(d) 的结果在下面

Counter({'mango': 1, 'pret': 1, 'orange': 1, 'préts': 1})

如何更改此代码以提供类似于下面的数据框 df2 的内容

d2 = {'s1': [3, 1], 's3':[2,1]}
df2 = pd.DataFrame(data=d2)

Answer 1

import pandas as pd
import re

d = {
  'Column_1': [u'mango pret Orange No manner', u' préts No scan '], 
  'Column_2': [u'read priority No', u'This is a priority noir '],
  'Column_3': [u'No add', u'yep']
}

df = pd.DataFrame(data=d)

list_1 = [u'Apple', u'Mango' ,u'Orange', u'pr[éeêè]t[s]?' ]
list_2 = [u'weather', u'r[ea]d' ,u'p[wr]iority', u'noir?' ]
list_3 = [u'n[eéè]d',u'snow[s]?', u'blanc?' ]

my_dict = {
  "s1": ['Column_1', list_1],
  "s2": ['Column_1', list_3],
  "s3": ['Column_2', list_2],
  "s4": ['Column_3', list_3],
  "s5": ['Column_2','Column_3',list_1]
}

d2 = dict()
for key, lst in my_dict.items():
  # Distinguish between columns and regex (assuming regex are stored in lists)
  col_names = filter(lambda x: isinstance(x, str), lst)
  regex_lists = filter(lambda x: isinstance(x, list), lst)
  # Concatenate all regex
  regex_list = reduce(lambda x, y: x+y, regex_lists)
  # For the considered columns, apply regex search in each cell and count
  map_function = lambda cell: len(re.findall(r'|'.join(regex_list).lower(), str(cell).lower()))
  df_regex_count = df[col_names].applymap(map_function)
  # Convert to desired output with lists to make a new dataframe
  d2[key] = map(sum, df_regex_count.values.tolist())

df2 = pd.DataFrame(data=d2)

输出：

    s1  s2  s3  s4  s5
0   3   0   1   0   0
1   1   0   2   0   0

请注意，s3 给出 [1, 2] 而不是 [2, 1]，因为 r[ea]d 没有捕获 read 和 noir? 捕获 noir。