[英]Checking # of times word occurs in text file (Python)
我在 python 工作,试图找到在文本文件中找到输入的次数,但 output 始终显示文本文件的字母数,而不是单词在文件中出现的次数。 我不确定是什么导致它变成 output 字母计数。
def occurrenceChecker(query):
wordInFile = 0
file = open("C:\Users\Noah\Desktop\1000words.txt")
documentText = file.read
for query in documentText():
wordInFile +=1
if wordInFile == 0:
print("Your argument occurred " + str(wordInFile) + " times, none.")
if wordInFile >= 1 and wordInFile <= 5:
print("Your argument occurred " + str(wordInFile) + " times, low.")
if wordInFile >= 6 and wordInFile <= 10:
print("Your argument occurred " + str(wordInFile) + " times, medium.")
if wordInFile >=11:
print("Your argument occurred " + str(wordInFile) + " times, high.")
Output:
Please enter an argument that you are searching for in the file: stuck
Your argument occurred 4875 times, high.
Your argument occurred 4875 times, high.
当您不想成为时,您正在呼叫 function,而当您想要成为时,您却没有呼叫 function。
for query in documentText():
for query in document_text:
将document_text = file.read
更改为document_text = file.read()
def occurrenceChecker(query):
word_in_file = 0
file = open("C:\Users\Noah\Desktop\1000words.txt")
document_text = file.read()
for query in document_text:
word_in_file +=1
if word_in_file == 0:
print("Your argument occurred " + str(word_in_file) + " times, none.")
if 1 <= word_in_file <= 5:
print("Your argument occurred " + str(word_in_file) + " times, low.")
if 6 <= word_in_file <= 10:
print("Your argument occurred " + str(word_in_file) + " times, medium.")
if word_in_file>=11:
print("Your argument occurred " + str(word_in_file) + " times, high.")
既然你已经有了,我们现在需要解决递增的问题。 代替
for query in document_text:
word_in_file +=1
你实际上想做这样的事情:
occurrences = document_text.count(query)
然后当然你应该有一个main()
并在所述main()
中调用你的 function
def main():
occurrenceChecker(input())
if __name__ == "__main__":
main()
资料来源: https://pythonexamples.org/python-count-occurrences-of-word-in-text-file/
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