如何从 CSV 文件创建字典

Question

我有一个像这样的 CSV 文件：

          w                                                syn
0     abaca   http://kaiko.getalp.org/dbnary/fra/Musa_textilis
1     abaca  http://kaiko.getalp.org/dbnary/fra/chanvre_de_...
2     abaca           http://kaiko.getalp.org/dbnary/fra/tagal
3     abaca   http://kaiko.getalp.org/dbnary/fra/Musa_textilis
4     abaca  http://kaiko.getalp.org/dbnary/fra/chanvre_de_...
..      ...                                                ...
95  abandon       http://kaiko.getalp.org/dbnary/fra/apostasie
96  abandon    http://kaiko.getalp.org/dbnary/fra/capitulation
97  abandon  http://kaiko.getalp.org/dbnary/fra/cession_de_...
98  abandon       http://kaiko.getalp.org/dbnary/fra/confiance
99  abandon       http://kaiko.getalp.org/dbnary/fra/défection

[100 rows x 2 columns]
6
{'abaca': 'tagal', 'abdomen': 'ventre', 'abricot': 'michemis', 'ADN': 'acide désoxyribonucléique', 'Indien': 'sauvage', 'abandon': 'défection'}

我试图创建一个字典，其中每个单词及其同义词。 我想出了这段代码，但最终的字典只包含该词的一个同义词，但正如您在 csv 文件中看到的那样，一个词可以有多个同义词。

# read specific columns of csv file using Pandas

df = pd.read_csv("sparql.csv", usecols = ["w","syn"]) #usecols = ["l","f","s","w","syn","synonyme"]
print(df)
liste_mot = df['w'].tolist()
liste_mot = set(liste_mot)
print(len(liste_mot))


liste_sys = []
dict_syn = {}


for index, row in df.iterrows():
    k, v = row
    sys = os.path.basename(v)
    if "_" in sys:
        sys = sys.split("_")
        sys = " ".join(sys)
        dict_syn[k] = sys
    else:
        dict_syn[k] = sys

print(dict_syn)

我想要得到的是每个单词作为关键字，并将所有同义词列表作为其值，但到目前为止，每个单词（w）我只能得到一个同义词（syn），而不是所有同义词。

Answer 1

这是一个部分基于您的代码的工作示例。 同义词放在一个列表中：

from io import StringIO

import pandas as pd


text = """
          w                                                syn
0     abaca   http://kaiko.getalp.org/dbnary/fra/Musa_textilis
1     abaca  http://kaiko.getalp.org/dbnary/fra/chanvre_de_...
2     abaca           http://kaiko.getalp.org/dbnary/fra/tagal
3     abaca   http://kaiko.getalp.org/dbnary/fra/Musa_textilis
4     abaca  http://kaiko.getalp.org/dbnary/fra/chanvre_de_...
95  abandon       http://kaiko.getalp.org/dbnary/fra/apostasie
95  abandon       http://kaiko.getalp.org/dbnary/fra/apostasie
96  abandon    http://kaiko.getalp.org/dbnary/fra/capitulation
97  abandon  http://kaiko.getalp.org/dbnary/fra/cession_de_...
98  abandon       http://kaiko.getalp.org/dbnary/fra/confiance
99  abandon       http://kaiko.getalp.org/dbnary/fra/défection
"""

# read in data
df = pd.read_csv(StringIO(text), sep='\s+')

# get the synonym out of the url
df['real_syn'] = df['syn'].str.extract('.*/(.*)')

# dictionary to write results to
result = {}

# loop over every row of the dataframe
for _, row in df[['w', 'real_syn']].iterrows():
    word = row['w']
    syn = row['real_syn']
   
    # check if word is already in result dictionary and make sure words are not added twice
    if result.get(word) and syn not in result[word]:
            result[word] = result[word] + [syn]
    else:
        # if word is not yet in dictionary, then add it a key, and add the synonym as a list
        result[word] = [syn] 
        
print(result)

Answer 2

我不确定您的 CSV 是否实际上是固定宽度的，或者这只是一个不错的打印输出。

如果您不需要 Pandas，Python 的标准 CSV 模块就可以胜任。

import csv
import os
import pprint

from collections import defaultdict

def syn_splitter(s):
    syn = os.path.basename(s)
    syn = syn.replace('_', ' ')
    return syn

# So we can just start appending syns, without having to "prime" the dictionary with an empty list
word_syn_map = defaultdict(list)

with open('sample.csv', 'r', newline='') as f:
    reader = csv.reader(f)
    next(reader)  # discard header

    for row in reader:
        w, syn = row
        syn = syn_splitter(syn)
        word_syn_map[w].append(syn)

pprint.pprint(word_syn_map)

# word_syn_map = dict(word_syn_map) if you want to get rid of the defaultdict wrapper

我模拟了sample.csv ：

w,syn
abaca,http://kaiko.getalp.org/dbnary/fra/Musa_textilis
abaca,http://kaiko.getalp.org/dbnary/fra/tagal
abaca,http://kaiko.getalp.org/dbnary/fra/Musa_textilis
abandon,http://kaiko.getalp.org/dbnary/fra/apostasie
abandon,http://kaiko.getalp.org/dbnary/fra/capitulation
abandon,http://kaiko.getalp.org/dbnary/fra/confiance
abandon,http://kaiko.getalp.org/dbnary/fra/défection

我得到了：

defaultdict(<class 'list'>,
            {'abaca': ['Musa textilis', 'tagal', 'Musa textilis'],
             'abandon': ['apostasie',
                         'capitulation',
                         'confiance',
                         'défection']})

Answer 3

另一种方法：

import os

df = pd.read_csv("sparql.csv", usecols=["w","syn"])
df["syn_new"] = df.syn.map(os.path.basename).str.replace("_", " ")
dict_syn = {
    key: group.syn_new.to_list()
    for key, group in df[["w", "syn_new"]].groupby("w")
}

您的样品的结果：

{'abaca': ['Musa textilis',
           'chanvre de ...',
           'tagal',
           'Musa textilis',
           'chanvre de ...'],
 'abandon': ['apostasie',
             'capitulation',
             'cession de ...',
             'confiance',
             'défection']}

你可以试试如果

df["syn_new"] = df.syn.str.rsplit("/", 1, expand=True)[1].str.replace("_", " ")

也可以，可能会更快。

也许您不想要list s 而是set s set为dict_syn值以避免重复：

...
    key: set(group.syn_new.to_list())
...