我有這個數(shù)據(jù)框： manufacturer description0 toyota toyota, gmc 10 years old.1 NaN gmc, Motor runs and drives good.2 NaN Motor old, in pieces.3 NaN 2 owner 0 rust. Cadillac.我想用從描述中獲取的關(guān)鍵字填充 NaN 值。為此，我創(chuàng)建了一個包含我想要的關(guān)鍵字的列表：keyword = ['gmc', 'toyota', 'cadillac']最后，我想循環(huán) DataFrame 中的每一行。將內(nèi)容從每行的“描述”列中拆分出來，如果該單詞也在“關(guān)鍵字”列表中，則將其添加到“制造商”列中。例如，它看起來像這樣： manufacturer description0 toyota toyota, gmc 10 years old.1 gmc gmc, Motor runs and drives good.2 NaN Motor old, in pieces.3 cadillac 2 owner 0 rust. Cadillac.感謝這個社區(qū)中的一位友好的人，我可以將我的代碼改進為：import rekeyword = ['gmc', 'toyota', 'cadillac']bag_of_words = []for i, description in enumerate(test3['description']):bag_of_words = re.findall(r"""[A-Za-z\-]+""", test3["description"][i])for word in bag_of_words: if word.lower() in keyword: test3.loc[i, 'manufacturer'] = word.lower()但我意識到第一行也改變了值，即使它不是 NaN： manufacturer description0 gmc toyota, gmc 10 years old.1 gmc gmc, Motor runs and drives good.2 NaN Motor old, in pieces.3 cadillac 2 owner 0 rust. Cadillac.我只想更改 NaN 值，但是當(dāng)我嘗試添加時：if word.lower() in keyword and test3.loc[i, 'manufacturer'] == np.nan:它沒有任何效果。
1 回答
智慧大石
TA貢獻1946條經(jīng)驗 獲得超3個贊
這是一個快速修復(fù)。你做錯了幾件事：
混淆了描述索引和描述本身（通過 解決
enumerate()）。bag_of_words應(yīng)該對每個單詞進行更新，而不是附加。正在迭代錯誤的項目（應(yīng)該是
word，不是bag_of_words）。
如果選擇直觀/常規(guī)的名稱，則可以很容易地看出一些錯誤。一定要花一些時間在這上面。
代碼：
from nltk.tokenize import RegexpTokenizer
# test3 = the main dataset
keyword = ['gmc', 'toyota', 'cadillac']
tokenizer = RegexpTokenizer('\w+|\$[\d\.]+|\S+')
for i, description in enumerate(test3['description']):
bag_of_words = tokenizer.tokenize(description.lower())
for word in bag_of_words:
if word in keyword:
test3.loc[i, 'manufacturer'] = word
輸出：
test3
Out[31]:
manufacturer description
0 toyota toyota, 10 years old.
1 gmc gmc, Motor runs and drives good.
2 NaN Motor old, in pieces.
3 cadillac 2 owner 0 rust. Cadillac.
通過 RegexpTokenizer 進行 re.findall()
我個人認為nltk是一個需要導(dǎo)入、安裝和部署的比較重的模塊。如果只進行字符串分割，我建議使用re.findall來提取有效的單詞模式。例如：
import re
# won't extract numbers, currency signs and apostrophes
re.findall(r"""[A-Za-z\-]+""", test3["description"][3])
# the output is much cleaner than before
Out[39]: ['owner', 'rust', 'Cadillac']
但這取決于用戶根據(jù)整個任務(wù)的選擇。
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