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wcfa.py

 1 # The authors of this work have released all rights to it and placed it
 2 # in the public domain under the Creative Commons CC0 1.0 waiver
 3 # (http://creativecommons.org/publicdomain/zero/1.0/).
 4 # 
 5 # THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
 6 # EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
 7 # MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
 8 # IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY
 9 # CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT,
10 # TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE
11 # SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
12 # 
13 # Retrieved from: http://en.literateprograms.org/Word_count_(Python,_functional)?oldid=18679
14 
15 from numpy import array, sum
16 # Gibbons' unlengths are nondecreasing runs, eg 00001111
17 drops=lambda bs: bs[1:] < bs[:-1]
18 blank=lambda cs: (cs==' ')+(cs=='\t')+(cs=='\n')
19 chars=lambda cs: array(tuple(cs))
20 words=lambda cs: drops(blank(chars('  '+cs)))
21 count=sum
22 wc=lambda s: count(words(s))
23 
24 if __name__=="__main__":
25         import sys
26         print wc(sys.stdin.read())


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