的Python解析出一個字符串的部分

Question

我有一個字符串，可以是這樣的：

方案1： 「some_field =（parameter_A-O-（8.869834109E-05））/（0.001 * 10）」

或

方案2： 「some_field =（parameter_A-（0.0005883943））/（0.001 * 10）」

如何可以在如下面的十進制格式解析出數字？

方案1： 第一個數字：-0.00008869834109 第二數：0.01

方案2： 第一個數字：0.0005883943 第二數：0.01

字符串格式保持不變，但數字格式和極性可以改變。

Answer 1

我認爲主要工作是提取真正包含所有周圍字符數字的棋子。這可以通過.split(),.find(),.rfind()和字符串內的索引來完成。

我的代碼假設只有一個等號'='，數字部分之間用'/'分隔，每個用圓括號括起來（如果多於括號級別，則在最裏面）可能是直接位於最內層支架左側的符號。如果有存儲在字符串多於兩個數字，因爲它抓住一個字符組的時間和附加所確定的值val到列表values

content = "some_field=(parameter_A-0-(8.869834109E-05))/(0.001*10)" 
content = "some_field=(parameter_A-(0.0005883943))/(0.001*10)" 
#or which way ever you give the character strings 

content = content.split('=') #split at equal sign 
if(len(content) != 2) : #check for exactly one equal sign 
    print ('something wrong in the data?') 

#in case information left of equal sign is needed 
fieldname = content[0] 

content = content[1] #now work on the part which is right of '=' 
content = content.split('/') 

values = [] 
for i in range(len(content)): 
    x = content[i] 

    pos_open = x.rfind('(') #find position of opening bracket '(', starting from right--> finds the right-most 
    pos_close = x.find(')') 
    #hence, the digits are in x[pos_open+1:pos_close] Check this by uncommenting the following line 
    #print(x[pos_open+1:pos_close] ) 

    #check whether there is a multiplication included in that part 
    if (x[pos_open+1:pos_close].find('*') < 0) : # .find() returns -1 if the character sequence is not found 
    val = float( x[pos_open+1:pos_close] ) # float() does the main work of conversion 
    else: 
    pos_dot = x[pos_open+1:pos_close].find('*') 
    factor1 = float( x[pos_open+1: pos_open+1 + pos_dot] ) 
    factor2 = float( x[pos_open+1 + pos_dot+1 : pos_close]) 
    val = factor1 * factor2 

    #check for negative sign: (sorry, your examples do not show clearly how the sign is specified) 
    if (pos_open > 0 and x[pos_open - 1] == '-'): # this checks the character before the bracket 
    #Note: in case of pos_open=0, the second part x[pos_open - 1] would look at the LAST character x[-1] 
    val = -val 

    values.append(val) 

#output 
print ('First: {0} Second: {1}' .format(values[0], values[1])) #standard way 
print ('First: ' + repr(values[0]) + ' Second: ' + repr(values[1])) #another standard way 

print ('First: {0:.12f} Second: {1:.7f}' .format(values[0], values[1])) # with format specified in two exemplary ways 

此代碼的提取部分也將起作用。