Hi I tried the following code to count the number of rows in a csv file :
import os
os.chdir('C:\\Users')
csv_file = open("count_rows.csv", "rb")
row_count = sum(1 for row in csv_file.read())
csv_file.close()
print (row_count)
The above code displays 18 as the result when the file has only 3 rows.
Any suggestions?
Thanks so much
The following line is not iterate lines, but each byte of the file content. (Iterating string object yields single-character strings)
row_count = sum(1 for row in csv_file.read())
To iterate lines, just iterate over the file object:
row_count = sum(1 for row in csv_file)
Here's a slighly modified version:
# Using `with` statement, you don't need to close manually.
# Use raw string literal: you can avoid escape
with open(r"C:\Users\count_rows.csv") as csv_file: # `r`: use text mode
row_count = sum(1 for row in csv_file)
print(row_count)
csv_file should be an iterator -- you can just run sum(1 for row in csv_file).
Also, it is best practice to open the file with a context manager:
with open('count_rows.csv') as buff:
row_count = sum(1 for _ in buff)
print(row_count)
any text file (incl. csv) can have lines counted as follows:
>>> fname = "file.txt"
>>> with open(fname) as f:
... for i, line in enumerate(f, 1):
... pass
... print i
enumerate generates numbers for all lines, assigning it to i.
When the loop terminates, i holds the number of lines. Note, that the enumerate has the 1
argument used, to set up numbering from 1. (thanks falsetru)
enumerate(f, 1), i will hold the number of lines. BTW, if the file is empty (0 byte), the last statement will raise NameError.
'rb'? That is for binary files. You just want'r'.