將數百萬和十億轉換爲Python列中的一個數字或整數

Question

我有一列值爲'10百萬'和'50億'，並且想要一個簡單的方法將其轉換爲數值來做一些更深入的分析。 我試圖

powers = {'billion': 10 ** 9, 'million': 10 ** 6} 

def f(s): 
    try: 
     power = s[-1] 
     return float(s[:-1]) * powers[power] 
    except TypeError: 
     return s 

df_2.applymap(f) 

更新：我的大熊貓柱由0（這是NaN的），幷包括百萬千萬的其他值。 我希望這比以前更清晰 我使用了下面@MobiusKlein推薦的方法。 所以這裏是有用的堆棧跟蹤錯誤。

--------------------------------------------------------------------------- 
TypeError         Traceback (most recent call last) 
<ipython-input-12-1db4b2353170> in <module>() 
    10  return float(quantity) * powers[magnitude] 
    11 
---> 12 df_2.applymap(f) 
    13 

/home/peadarcoyle/.virtualenvs/Ipython/local/lib/python2.7/site-packages/pandas/core/frame.pyc in applymap(self, func) 
    3725     x = lib.map_infer(_values_from_object(x), f) 
    3726    return lib.map_infer(_values_from_object(x), func) 
-> 3727   return self.apply(infer) 
    3728 
    3729  #---------------------------------------------------------------------- 

/home/peadarcoyle/.virtualenvs/Ipython/local/lib/python2.7/site-packages/pandas/core/frame.pyc in apply(self, func, axis, broadcast, raw, reduce, args, **kwds) 
    3556      if reduce is None: 
    3557       reduce = True 
-> 3558      return self._apply_standard(f, axis, reduce=reduce) 
    3559    else: 
    3560     return self._apply_broadcast(f, axis) 

/home/peadarcoyle/.virtualenvs/Ipython/local/lib/python2.7/site-packages/pandas/core/frame.pyc in _apply_standard(self, func, axis, ignore_failures, reduce) 
    3646    try: 
    3647     for i, v in enumerate(series_gen): 
-> 3648      results[i] = func(v) 
    3649      keys.append(v.name) 
    3650    except Exception as e: 

/home/peadarcoyle/.virtualenvs/Ipython/local/lib/python2.7/site-packages/pandas/core/frame.pyc in infer(x) 
    3724     f = com.i8_boxer(x) 
    3725     x = lib.map_infer(_values_from_object(x), f) 
-> 3726    return lib.map_infer(_values_from_object(x), func) 
    3727   return self.apply(infer) 
    3728 

/home/peadarcoyle/.virtualenvs/Ipython/local/lib/python2.7/site-packages/pandas/lib.so in pandas.lib.map_infer (pandas/lib.c:56671)() 

<ipython-input-12-1db4b2353170> in f(num_str) 
     4 
     5 def f(num_str): 
----> 6 match = re.search(r"([0-9\.]+)\s?(million|billion)", num_str) 
     7 if match is not None: 
     8  quantity = match.group(0) 

/home/peadarcoyle/.virtualenvs/Ipython/lib/python2.7/re.pyc in search(pattern, string, flags) 
    140  """Scan through string looking for a match to the pattern, returning 
    141  a match object, or None if no match was found.""" 
--> 142  return _compile(pattern, flags).search(string) 
    143 
    144 def sub(pattern, repl, string, count=0, flags=0): 

TypeError: ('expected string or buffer', u'occurred at index Intended_Investment') 

Answer 1

您的函數查詢字符串的數字不考慮空格或超過一位數字領先。嘗試一些更復雜的：

import re 
powers = {'billion': 10 ** 9, 'million': 10 ** 6} 

def f(num_str): 
    match = re.search(r"([0-9\.]+)\s?(million|billion)", num_str) 
    if match is not None: 
     quantity = match.group(0) 
     magnitude = match.group(1) 
     return float(quantity) * powers[magnitude] 

這將引發錯誤，如果它不能提取字符串正確的令牌，但它處理的白色空間和權力的不規則的數量。如果您擔心浮點錯誤，請使用int（）將數量轉換爲數字類型，但請確保不處理小數。如果你是的話，你可以用大小來玩遊戲來解決這個問題，但是它會讓代碼比第一遍更復雜。