Question

我需要知道给定字符出现在字符串中的次数。

该解决方案应该能够在大多数与POSIX兼容的shell解释器下运行，并且应该具有尽可能少的依赖关系。

Answer 1

这是一个仅依赖于import os import win32com.client import numpy as np from utils import DATA_DIR def read_from_excel(): place=os.path.join(DATA_DIR,'book1.xlsx') Excel=win32com.client.Dispatch('Excel.Application') wb=Excel.Workbooks.Open(place) sheet=wb.ActiveSheet TI3_name=[r[0].value for r in sheet.Range("Table[TI3_Name]")] TI4_name=[r[0].value for r in sheet.Range("Table[TI4_Name]")] TI5_name=[r[0].value for r in sheet.Range("Table[TI5_Name]")] Frank_name=[r[0].value for r in sheet.Range("Table[Frank_name]")] Shanh_name=[r[0].value for r in sheet.Range("Table[Shanh_name]")] Manil_name=[r[0].value for r in sheet.Range("Table[Manil_name]")] TI3_value=[r[0].value for r in sheet.Range("Table[TI3]")] TI4_value=[r[0].value for r in sheet.Range("Table[TI4]")] TI5_value=[r[0].value for r in sheet.Range("Table[TI5]")] Frank_value=[r[0].value for r in sheet.Range("Table[Frank]")] Shanh_value=[r[0].value for r in sheet.Range("Table[Shanh]")] Manil_value=[r[0].value for r in sheet.Range("Table[Manil]")] TI3_name=[x for x in TI3_name if str(x)!='None'] TI4_name=[x for x in TI4_name if str(x)!='None'] TI5_name=[x for x in TI5_name if str(x)!='None'] Frank_name=[x for x in Frank_name if str(x)!='None'] Shanh_name=[x for x in Shanh_name if str(x)!='None'] Manil_name=[x for x in Manil_name if str(x)!='None'] TI3_value=[x for x in TI3_value if str(x)!='None'] TI4_value=[x for x in TI4_value if str(x)!='None'] TI5_value=[x for x in TI5_value if str(x)!='None'] Frank_value=[x for x in Frank_value if str(x)!='None'] Shanh_value=[x for x in Shanh_value if str(x)!='None'] Manil_value=[x for x in Manil_value if str(x)!='None'] TI3=list(zip(TI3_name,TI3_value)) TI4=list(zip(TI4_name, TI4_value)) TI5=list(zip(TI5_name, TI5_value)) Frank=list(zip(Frank_name, Frank_value)) Shanh=list(zip(Shanh_name, Shanh_value)) Manil=list(zip(Manil_name, Manil_value)) TI3=np.array(TI3) TI4=np.array(TI4) TI5=np.array(TI5) Frank=np.array(Frank) Shanh=np.array(Shanh) Manil=np.array(Manil) return TI3, TI4, TI5, Frank, Shanh, Manil def find_number_uniq_heroes(TI3, TI4, TI5, Frank, Shanh, Manil): x1=TI3[:,0] x2=TI4[:,0] x3=TI5[:,0] x4=Frank[:,0] x5=Shanh[:,0] x6=Manil[:,0] x=np.concatenate((x1,x2,x3,x4,x5,x6),0) y=np.unique(x) return y def number_popular(x,y,hero): try: i=x.tolist().index(hero) return float(y[i]) except ValueError: return float(0) def count_popularity(uniq_heroes,TI3, TI4, TI5, Frank, Shanh, Manil): n=len(uniq_heroes) popul=np.zeros(n) x1=TI3[:,0] x2=TI4[:,0] x3=TI5[:,0] x4=Frank[:,0] x5=Shanh[:,0] x6=Manil[:,0] y1=TI3[:,1] y2=TI4[:,1] y3=TI5[:,1] y4=Frank[:,1] y5=Shanh[:,1] y6=Manil[:,1] i=0 for hero in uniq_heroes: pop=float(0) pop=number_popular(x1,y1,hero) pop=pop+number_popular(x2,y2,hero) pop=pop+number_popular(x3,y3,hero) pop=pop+number_popular(x4,y4,hero) pop=pop+number_popular(x5,y5,hero) pop=pop+number_popular(x6,y6,hero) popul[i]=float(pop) i=i+1 return list(zip(uniq_heroes,popul)) def sort_mass(x): x=sorted(x,key=lambda x_entry:x_entry[1],reverse=True) return x TI3, TI4, TI5, Frank, Shanh, Manil = read_from_excel() uniq_heroes=find_number_uniq_heroes(TI3, TI4, TI5, Frank, Shanh, Manil) un_her_with_popul=count_popularity(uniq_heroes,TI3, TI4, TI5, Frank, Shanh, Manil) sorted_mas=sort_mass(un_her_with_popul) print(sorted_mas[:10])，echo，tr和test的shell函数：

${#

用法示例：

charcount() {
  local char="$1"; shift;
  result="$(echo "$*" | tr -cd "$char")"; result=${#result};
  test $result -gt 0 && echo $result
}

这是一个更具体的解决方案，只能测试字符串中是否存在字符。

# charcount 'a' 'test' || echo 'char not found'
char not found

# charcount 'a' 'a test a' && echo 'found'
2
found

Shell脚本：计算给定字符在字符串中出现的次数

1 个答案: