我需要在串列中每行 id 的開頭添加五次。
file_id像
5
8
6
9
text_file像
pla pla pla
text text text
dsfdfdfdfd
klfdklfkdkf
poepwoepwo
lewepwlew
結果應該是
5 pla pla pla
5 text text text
5 dsfdfdfdfd
5 klfdklfkdkf
5 poepwoepwo
8 lewepwlew
依此類推.. id 的數量等于 5000 個 id，文本等于 25000 個句子 .. 每個 id 將包含五個句子。我試圖做這樣的事情
import fileinput
import sys
f = open("id","r")
List=[]
for id in f.readlines():
List.append(id)
file_name = 'text.txt'
with open(file_name,'r') as fnr:
text = fnr.readlines()
i=0
text = "".join([List[i] " " line.rstrip() for line in text])
with open(file_name,'w') as fnw:
fnw.write(text)
但得到了結果
5
pla pla pla5
text text text5
dsfdfdfdfd5
klfdklfkdkf5
poepwoepwo5
lewepwlew5
uj5u.com熱心網友回復：
你可以試試這個：雖然我還沒有測驗過
lst = []
with open("id","r") as f:
ids = f.read().split('\n')
file_name = 'text.txt'
with open(file_name,'r') as fnr:
text = fnr.read().split('\n')
counter = 0
for id_num in ids:
for _ in range(5):
if counter >= len(text):
break
lst.append(id_num " " text[counter])
counter = 1
text = '\n'.join(lst)
with open(file_name,'w') as fnw:
fnw.write(text)
輸出：
1 pla pla pla
1 text text text
1 dsfdfdfdfd
1 klfdklfkdkf
1 poepwoepwo
2 lewepwlew
2
uj5u.com熱心網友回復：
代替：
i=0
text = "".join([List[i] " " line.rstrip() for line in text])
你可以試試：
new_text = []
for i, line in enumerate(text):
new_text.append(List[i % 5] " " line.rstrip())
然后寫入new_text檔案。
轉載請註明出處，本文鏈接：https://www.uj5u.com/net/478182.html