熊猫：将列强制转换为字符串不起作用

熊猫：将列强制转换为字符串不起作用

dtype

的

string

，

dict

，

list

始终是

object

，测试

type

需要选择列例如，由一些价值

iat

：

type(resultstatsDF['file'].iat[0])

样品：

resultstatsDF = pd.Dataframe({'file':['a','d','f']})print (resultstatsDF)  file0    a1    d2    fprint (type(resultstatsDF['file'].iloc[0]))<class 'str'>print (resultstatsDF['file'].apply(type))0    <class 'str'>1    <class 'str'>2    <class 'str'>Name: file, dtype: object

样品：

df = pd.Dataframe({'strings':['a','d','f'],        'dicts':[{'a':4}, {'c':8}, {'e':9}],        'lists':[[4,8],[7,8],[3]],        'tuples':[(4,8),(7,8),(3,)],        'sets':[set([1,8]), set([7,3]), set([0,1])] })print (df)      dicts   lists    sets strings  tuples0  {'a': 4}  [4, 8]  {8, 1}       a  (4, 8)1  {'c': 8}  [7, 8]  {3, 7}       d  (7, 8)2  {'e': 9}     [3]  {0, 1}       f    (3,)

所有值都相同

dtypes

：

print (df.dtypes)dicts      objectlists      objectsets       objectstrings    objecttuples     objectdtype: object

但是

type

不同，如果需要循环检查：

for col in df:    print (df[col].apply(type))0    <class 'dict'>1    <class 'dict'>2    <class 'dict'>Name: dicts, dtype: object0    <class 'list'>1    <class 'list'>2    <class 'list'>Name: lists, dtype: object0    <class 'set'>1    <class 'set'>2    <class 'set'>Name: sets, dtype: object0    <class 'str'>1    <class 'str'>2    <class 'str'>Name: strings, dtype: object0    <class 'tuple'>1    <class 'tuple'>2    <class 'tuple'>Name: tuples, dtype: object

或列的第一个值：

print (type(df['strings'].iat[0]))<class 'str'>print (type(df['dicts'].iat[0]))<class 'dict'>print (type(df['lists'].iat[0]))<class 'list'>print (type(df['tuples'].iat[0]))<class 'tuple'>print (type(df['sets'].iat[0]))<class 'set'>

与

boolean indexing

如果可能混合柱（然后一些熊猫功能可以分成）是可能的过滤器通过

type

：

df = pd.Dataframe({'mixed':['3', 5, 9,'2']})print (df)  mixed0     31     52     93     2print (df.dtypes)mixed    objectdtype: object

---

for col in df:    print (df[col].apply(type))0    <class 'str'>1    <class 'int'>2    <class 'int'>3    <class 'str'>Name: mixed, dtype: object#python 3 - string#python 2 - basestringmask = df['mixed'].apply(lambda x: isinstance(x,str))print (mask)0     True1    False2    False3     TrueName: mixed, dtype: booldf = df[mask]print (df)  mixed0     33     2