我正在尝试在数据框中搜索字典值中列出的某些单词(如果存在),将其替换为值的键。
units_dic= {'grams':['g','Grams'],
'kg' :['kilogram','kilograms']}
问题是某些单位缩写是字母,所以它也会替换所有字母,我只想在替换字母后加上数字以确保它是一个单位。
数据框
Id | test
---------
1 |'A small paperclip has a mass of about 111 g'
2 |'1 kilogram =1000 g'
3 |'g is the 7th letter in the ISO basic Latin alphabet'
替换循环
x = df.copy()
for k in units_dic:
for i in range(len(x['test'])):
for w in units_dic[k]:
x['test'][i] = str(x['test'][i]).replace(str(w), str(k))
输出
Id | test
---------
1 |'A small paperclip has a mass of about 111 grams'
2 |'1 kg =1000 grams'
3 |'grams is the 7th letter in the ISO basic Latin alphabet'
答案 0 :(得分:1)
尝试:
for key, val in units_dic.items():
df['test'] = df['test'].replace("\d+[ ]*" + "|".join(val) , key , regex=True)
答案 1 :(得分:1)
正则表达式可用于翻转字典。
import re
d = {i: k for k, v in units_dic.items() for i in v}
u = r'|'.join(d)
v = fr'(\d+\s?)\b({u})\b'
df.assign(test=[re.sub(v, lambda x: x.group(1) + d[x.group(2)], el) for el in df.test])
Id test
0 1 A small paperclip has a mass of about 111 grams
1 2 1 kg =1000 grams
2 3 g is the 7th letter in the ISO basic Latin alp...
答案 2 :(得分:0)
我们可以在此处使用var data = [{ Sum: 4580, class: "01", pID: 1 }, { Sum: 580, class: "01", pID: 2 }, { Sum: 1280, class: "01", pID: 3 }, { Sum: 5580, class: "02", pID: 1 }, { Sum: 280, class: "02", pID: 2 }, { Sum: 380, class: "02", pID: 3 }],
result = Array.from(data
.reduce(
(m, o) => m.set(o.class, Object.assign(m.get(o.class) || {}, { class: o.class, ['pID' + o.pID]: o.Sum })),
new Map
)
.values()
);
console.log(result);
的{{1}}功能,该功能可以指定它前面必须有一个数字和 optional 一个空格:>
.as-console-wrapper { max-height: 100% !important; top: 0; }
lookbehind
说明
首先,我们使用raw + fstring:regex
正则表达式:
for k, v in units_dic.items():
df['test'] = df['test'].str.replace(f"(?<=[0-9])\s*({'|'.join(v)})\b", f' {k}')
=后跟数字print(df)
Id test
0 1 'A small paperclip has a mass of about 111 grams'
1 2 '1 kg =1000 grams'
2 3 'g is the 7th letter in the ISO basic Latin al...
是空格fr'sometext'
给我们您字典中以?<=[0-9]
分隔的值
是正则表达式中的\s*
运算符