我试图找到这三个人口最多的城市部件(BYDEL)于1992年Python的分析csv文件
我有一个CSV文件看起来像这样:http://data.kk.dk/dataset/9070067f-ab57-41cd-913e-bc37bfaf9acd/resource/9fbab4aa-1ee0-4d25-b2b4-b7b63537d2ec/download/befkbhalderkoencivst.csv>
CSV文件可以作为解释:
AAR:在观测在哪一年
BYDEL:哪个城市的一部分,由一个整数描述我所含n遵循字典; 1 = Indre By,2 =Østerbro,3 =Nørrebro,4 = Vesterbro/Kgs。 Enghave,5 = Valby,6 =Vanløse,7 =Brønshøj-苏姆,8 = Bispebjerg,9 =阿迈厄OST,10 =阿迈厄背心,99 = Udenfor inddeling
ALDER:所观察到的人的年龄
PERSONER:与该行
我有一个解决方案,但它是非常重复,我认为它可以做更聪明,但我没有与蟒蛇足够的经验给定功能的意见数量。任何人都可以将我指向正确的方向吗?
我的代码/解决方案是这样的:
df = pd.read_csv('befkbh.csv',quotechar='"',skipinitialspace=True, delimiter=',', encoding='latin1').fillna(0)
data = df.as_matrix()
Q31 = collections.defaultdict(list)
Q32 = collections.defaultdict(list)
Q33 = collections.defaultdict(list)
Q34 = collections.defaultdict(list)
Q35 = collections.defaultdict(list)
Q36 = collections.defaultdict(list)
Q37 = collections.defaultdict(list)
Q38 = collections.defaultdict(list)
Q39 = collections.defaultdict(list)
Q310 = collections.defaultdict(list)
Q399 = collections.defaultdict(list)
for row in data:
key = row[0]
if key == "" or key == 0: continue
if key == 1992:
if row[2] == 1:
val = 0 if(row[5]) =="" else float(row[5])
Q31.setdefault(key,[]).append(val)
if row[2] == 2:
val = 0 if(row[5]) =="" else float(row[5])
Q32.setdefault(key,[]).append(val)
if row[2] == 3:
val = 0 if(row[5]) =="" else float(row[5])
Q33.setdefault(key,[]).append(val)
if row[2] == 4:
val = 0 if(row[5]) =="" else float(row[5])
Q34.setdefault(key,[]).append(val)
if row[2] == 5:
val = 0 if(row[5]) =="" else float(row[5])
Q35.setdefault(key,[]).append(val)
if row[2] == 6:
val = 0 if(row[5]) =="" else float(row[5])
Q36.setdefault(key,[]).append(val)
if row[2] == 7:
val = 0 if(row[5]) =="" else float(row[5])
Q37.setdefault(key,[]).append(val)
if row[2] == 8:
val = 0 if(row[5]) =="" else float(row[5])
Q38.setdefault(key,[]).append(val)
if row[2] == 9:
val = 0 if(row[5]) =="" else float(row[5])
Q39.setdefault(key,[]).append(val)
if row[2] == 10:
val = 0 if(row[5]) =="" else float(row[5])
Q310.setdefault(key,[]).append(val)
if row[2] == 99:
val = 0 if(row[5]) =="" else float(row[5])
Q399.setdefault(key,[]).append(val)
Q312 = {}
for k, v in Q31.items(): Q312[k] = sum(v)
for k, v in Q312.items(): print ("{}:{}".format(k,v))
Q322 = {}
for k, v in Q32.items(): Q322[k] = sum(v)
for k, v in Q322.items(): print ("{}:{}".format(k,v))
Q332 = {}
for k, v in Q33.items(): Q332[k] = sum(v)
for k, v in Q332.items(): print ("{}:{}".format(k,v))
Q342 = {}
for k, v in Q34.items(): Q342[k] = sum(v)
for k, v in Q342.items(): print ("{}:{}".format(k,v))
Q352 = {}
for k, v in Q35.items(): Q352[k] = sum(v)
for k, v in Q352.items(): print ("{}:{}".format(k,v))
Q362 = {}
for k, v in Q36.items(): Q362[k] = sum(v)
for k, v in Q362.items(): print ("{}:{}".format(k,v))
Q372 = {}
for k, v in Q37.items(): Q372[k] = sum(v)
for k, v in Q372.items(): print ("{}:{}".format(k,v))
Q382 = {}
for k, v in Q38.items(): Q382[k] = sum(v)
for k, v in Q382.items(): print ("{}:{}".format(k,v))
Q392 = {}
for k, v in Q39.items(): Q392[k] = sum(v)
for k, v in Q392.items(): print ("{}:{}".format(k,v))
Q3102 = {}
for k, v in Q310.items(): Q3102[k] = sum(v)
for k, v in Q3102.items(): print ("{}:{}".format(k,v))
Q3992 = {}
for k, v in Q399.items(): Q3992[k] = sum(v)
for k, v in Q3992.items(): print ("{}:{}".format(k,v))
究竟那种单挑我一直在寻找。谢谢大家,熊猫将在未来为我节省大量的时间和精力:) – Rainoa
完美的答案,很好 – single430
我试图让所有不同的年份和pd.unique()的前3名只给我所有不同年份的数组。什么是最聪明的熊猫方式呢?先谢谢了! – Rainoa