python如何统计文件行数 python统计代码行数的几个例子

#!/usr/bin/python
#

import time
def block(file,size=65536):
    while True:
        nb = file.read(size)
        if not nb:
           break
        yield nb
def getLineCount(filename):
    with open(filename,"r",encoding="utf-8") as f:
        return sum(line.count("n") for line in block(f))
if __name__ == "__main__":
    import sys
    import os
    if len(sys.argv) != 2:
        print("error imput argument")
        sys.exit(-1)
    if not os.path.isfile(sys.argv[1]) :
       print(sys.argv + " is not a file")
       sys.exit(-1)
    start_time = time.time()
    print(getLineCount(sys.argv[1]))
    print(time.time() - start_time ,"seconds")

#!/usr/bin/python
'''
        File      : count.py
        Author    : Mike
        E-Mail    : Mike_Zhang@live.com
'''
import sys,os
extens = [".c",".cpp",".hpp",".h"]
linesCount = 0
filesCount = 0
def funCount(dirName):
    global extens,linesCount,filesCount
    for root,dirs,fileNames in os.walk(dirName):
        for f in fileNames:
            fname = os.path.join(root,f)
            try :
                ext = f[f.rindex('.'):]
                if(extens.count(ext) > 0):
                    print 'support'
                    filesCount += 1
                    print fname
                    l_count = len(open(fname).readlines())
                    print fname," : ",l_count
                    linesCount += l_count
                else:
                    print ext," : not support"
            except:
                print "Error occur!"
                pass

if len(sys.argv) > 1 :
    for m_dir in sys.argv[1:]:       
        print m_dir
        funCount(m_dir)
else :
    funCount(".")       

print "files count : ",filesCount
print "lines count : ",linesCount
raw_input("Press Enter to continue")

in recent Python 2.7, we have new OrderedDict type, which remembers the order in which the items were added.
>>> d = {"third": 3, "first": 1, "fourth": 4, "second": 2}
>>> for k, v in d.items():
...     print "%s: %s" % (k, v)
...
second: 2
fourth: 4
third: 3
first: 1
>>> d
{'second': 2, 'fourth': 4, 'third': 3, 'first': 1}To make a new ordered dictionary from the original, sorting by the values:
>>> from collections import OrderedDict
>>> d_sorted_by_value = OrderedDict(sorted(d.items(), key=lambda x: x[1]))The OrderedDict behaves like a normal dict:
>>> for k, v in d_sorted_by_value.items():
...     print "%s: %s" % (k, v)
...
first: 1
second: 2
third: 3
fourth: 4
>>> d_sorted_by_value
OrderedDict([('first': 1), ('second': 2), ('third': 3), ('fourth': 4)])

#coding=utf-8
import operator
f = open("f.txt")
count_dict = {}
for line in f.readlines():
    line = line.strip()
    count = count_dict.setdefault(line, 0)
    count += 1
    count_dict[line] = count
sorted_count_dict = sorted(count_dict.iteritems(), key=operator.itemgetter(1), reverse=True)
for item in sorted_count_dict:
    print "%s,%d" % (item[0], item[1])

>>> help(sorted)
Help on built-in function sorted in module __builtin__:
sorted(...)
    sorted(iterable, cmp=None, key=None, reverse=False) --> new sorted list