[英]how can I set the string in pandas data frame in the same position in all rows?
我正在嘗試使用熊貓拆分字符串以構成CPC層次結構。 這是我的數據框df_all_new_p
CPC
0 Y10T403/4602
1 H02S20/00
2 H01L31/02168
我正在考慮將Lv.1包含一個字母,Lv2包含兩個字母,Lv3包含2-3個字母,Lv4,5,6,7,8 ..作為后面的字母,使CPC達到6-10的水平'/'
例如。
Y10T403/4602 -> Y, 10, T, 403, 43/4, 43/46, 43/460, 43/4602
H02S20/00 -> H, 02, S, 20, 20/0, 20/00
H01L31/02168-> H, 01, L, 31, 31/0, 31/02, 31/021, 31/0216, 31/02168
但是當我運行我的代碼時
if df_all_new_p['CPC'].map(lambda x: x[0:7]).str.contains('/').any():
df_all_new_p['Lv1'] = df_all_new_p['CPC'].map(lambda x: x[0:1])
df_all_new_p['Lv2'] = df_all_new_p['CPC'].map(lambda x: x[1:3])
df_all_new_p['Lv3'] = df_all_new_p['CPC'].map(lambda x: x[3:4])
df_all_new_p['Lv4'] = df_all_new_p['CPC'].map(lambda x: x[4:6])
df_all_new_p['Lv5'] = df_all_new_p['CPC'].map(lambda x: x[4:8])
df_all_new_p['Lv6'] = df_all_new_p['CPC'].map(lambda x: x[4:9])
elif df_all_new_p['CPC'].map(lambda x: x[0:8]).str.contains('/').any():
df_all_new_p['Lv1'] = df_all_new_p['CPC'].map(lambda x: x[0:1])
df_all_new_p['Lv2'] = df_all_new_p['CPC'].map(lambda x: x[1:3])
df_all_new_p['Lv3'] = df_all_new_p['CPC'].map(lambda x: x[3:4])
df_all_new_p['Lv4'] = df_all_new_p['CPC'].map(lambda x: x[4:7])
df_all_new_p['Lv5'] = df_all_new_p['CPC'].map(lambda x: x[7:9])
df_all_new_p['Lv6'] = df_all_new_p['CPC'].map(lambda x: x[7:10])
else:
df_all_new_p['Lv1'] = df_all_new_p['CPC'].map(lambda x: x[0:1])
df_all_new_p['Lv2'] = df_all_new_p['CPC'].map(lambda x: x[1:3])
df_all_new_p['Lv3'] = df_all_new_p['CPC'].map(lambda x: x[3:4])
df_all_new_p['Lv4'] = df_all_new_p['CPC'].map(lambda x: x[4:8])
df_all_new_p['Lv5'] = df_all_new_p['CPC'].map(lambda x: x[8:10])
df_all_new_p['Lv6'] = df_all_new_p['CPC'].map(lambda x: x[8:11])
df_all_new_p
CPC Lv1 Lv2 Lv3 Lv4 Lv5 Lv6
0 Y10T403/4602 Y1 0 T4 03 4602
1 H02S20/00 H 02 S 20 20/0 20/00
2 H01L31/02168 H0 1 L3 1/ 02168
結果表明,只有1 H02S20/00
得到正確答案,而另一行得到錯誤結果。 我認為原因是由每一行中的字符位置引起的。 因此,我想知道有什么方法可以解決這個問題嗎?
我希望看到這樣的結果。
CPC 1 2 3 4 5 6
Y10T403/4602 Y 10 T 403 43/4 43/46
H02S20/00 H 02 S 20 20/0 20/00
H01L31/02168 H 01 L 31 31/0 31/02
先說一下,也許有更有效的方法可以做到這一點。 也就是說,您可以使用str.find('/')來幫助建立索引:
df=pd.DataFrame({'a':[1,2,3],'CPC':['Y10T403/4602','H02S20/00','H01L31/02168']})
a CPC
0 1 Y10T403/4602
1 2 H02S20/00
2 3 H01L31/02168
[i[i.find('/')-2:i.find('/')+3] for i in df['CPC']]
['03/46', '20/00', '31/02']
因此,您可以定義一個函數以傳遞給df.apply()
def parse_cpc(val):
elems=[]
elems.append(val[0])
elems.append(val[1:3])
elems.append(val[3])
elems.append(val[4:val.find('/')])
elems.append(val[val.find('/')-2:val.find('/')+2])
elems.append(val[val.find('/')-2:val.find('/')+3])
return elems
並應用它,然后將該列拆分為多個列(*編輯以刪除不必要的lambda)
df['p']=df['CPC'].apply(parse_cpc)*
a CPC p
0 1 Y10T403/4602 [Y, 10, T, 403, 03/4, 03/46]
1 2 H02S20/00 [H, 02, S, 20, 20/0, 20/00]
2 3 H01L31/02168 [H, 01, L, 31, 31/0, 31/02]
df[[1,2,3,4,5,6]]=pd.DataFrame(df['p'].values.tolist())
a CPC p 1 2 3 4 5 6
0 1 Y10T403/4602 [Y, 10, T, 403, 03/4, 03/46] Y 10 T 403 03/4 03/46
1 2 H02S20/00 [H, 02, S, 20, 20/0, 20/00] H 02 S 20 20/0 20/00
2 3 H01L31/02168 [H, 01, L, 31, 31/0, 31/02] H 01 L 31 31/0 31/02
然后刪除過渡列
df.drop('p', axis=1, inplace=True)
a CPC 1 2 3 4 5 6
0 1 Y10T403/4602 Y 10 T 403 03/4 03/46
1 2 H02S20/00 H 02 S 20 20/0 20/00
2 3 H01L31/02168 H 01 L 31 31/0 31/02
這是在自定義函數pandas.concat
regex
模式與Series.extract
和pandas.concat
一起使用的另一種可能方法:
def cpc_hierarchy(cpc_series):
pat1 = '(.)(\d{2})(.)(\d{2,3})'
pat2 = '(\d{2}\/\d{1})'
pat3 = '(\d{2}\/\d{2})'
expanded = (pd.concat([
cpc_series.to_frame(),
cpc_series.str.extract(pat1),
cpc_series.str.extract(pat2),
cpc_series.str.extract(pat3)],
ignore_index=True,
axis=1).set_index(0).rename_axis('CPC', 0))
return expanded
print(cpc_hierarchy(df['CPC']))
1 2 3 4 5 6
CPC
Y10T403/4602 Y 10 T 403 03/4 03/46
H02S20/00 H 02 S 20 20/0 20/00
H01L31/02168 H 01 L 31 31/0 31/02
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