blocks|key|2195008|text|是的，这是可能的。您使用dtype|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2195009|import+pandas+as+pd
import+numpy+as+np

df+=+pd.DataFrame({'a':+['a','b','c','d']})
if+df['a'].dtype+!=+np.number:
++++print('yes')
else:
++++print('no')|code-block|syntax|javascript|2195010|还可以使用select_dtypes按dtype选择列|2195011|df_subset+=+df.select_dtypes(exclude=[np.number])
#+Now+apply+you+can+label+encode+your+df_subset|2195012|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dtype.html|1|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.select_dtypes.html^0|C|5|C|5|0|0|0|5|D|5|D|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@$9|18|A|19|1|1A]]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

Yes, its possible. You use <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dtype.html" rel="noreferrer"><code>dtype</code></a>

<pre><code>import pandas as pd
import numpy as np

df = pd.DataFrame({'a': ['a','b','c','d']})
if df['a'].dtype != np.number:
 print('yes')
else:
 print('no')
</code></pre>

You can also select your columns by dtype using <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.select_dtypes.html" rel="noreferrer"><code>select_dtypes</code></a>

<pre><code>df_subset = df.select_dtypes(exclude=[np.number])
# Now apply you can label encode your df_subset
</code></pre>

blocks|key|1714905|text|Scratch‘N’answer的答案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1714906|>>>+df+=+pd.DataFrame({'a':+['a','b','c','d'],+'b':+[1,+'b',+'c',+2],+'c':+[np.nan,+2,+3,+4]})
>>>+df+
+++a++b++++c
0++a++1++NaN
1++b++b++2.0
2++c++c++3.0
3++d++2++4.0

>>>+dict(filter(lambda+x:+x[1]+!=+np.number,+list(zip(df.columns,+df.dtypes))))
{'a':+dtype('O'),+'b':+dtype('O')}|code-block|syntax|javascript|1714907|所以我添加了一些混合类型的列。您可以看到，filter+%2B+dict方法生成了哪些列的数据类型超出np.number界限的键:值映射。这在规模上应该很好地发挥作用。您还可以尝试将每一列强制为特定类型(例如int)，然后在无法将字符串列转换为int时捕获ValueError异常。有很多方法可以做到这一点。|offset|length|style|CODE|1714908|entityMap^0|0|0|L|6|U|4|1D|9|2U|3|3C|3|3I|A|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|N|$]]

Expanding on Scratch'N'Purr's answer:

<pre><code>&gt;&gt;&gt; df = pd.DataFrame({'a': ['a','b','c','d'], 'b': [1, 'b', 'c', 2], 'c': [np.nan, 2, 3, 4]})
&gt;&gt;&gt; df 
 a b c
0 a 1 NaN
1 b b 2.0
2 c c 3.0
3 d 2 4.0

&gt;&gt;&gt; dict(filter(lambda x: x[1] != np.number, list(zip(df.columns, df.dtypes))))
{'a': dtype('O'), 'b': dtype('O')}
</code></pre>

So I've added some columns with mixed types. You can see that the <code>filter</code> + <code>dict</code> approach yields key: value mappings of which columns have dtypes outside of the bounds of <code>np.number</code>. This ought to work well at scale. You could also try coercing each column to a specific type (e.g. <code>int</code>) and then catch the <code>ValueError</code> exception when you can't convert a string column to <code>int</code>. Lots of ways to do this.

blocks|key|2195092|text|请注意，上面的答案将包括DateTime、TimeStamp、Category和其他数据类型。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2195093|使用object有更多的限制(尽管我不确定其他dtypes是否也使用object数据类型)：|offset|length|style|CODE|2195094|2195095|创建数据帧：|ordered-list-item|2195096|2195097|df+=+pd.DataFrame({+'a'：'A'，'B'，'c'，'d'，'b'：1，'B'，'c'，2，'c'：np.nan，2，3，4，'d'：'A'，'b'，'b'，'a'，'e'：今日(‘df'd'.astype('category')’)})df‘=pd.to_datetime|2195098|它看起来像这样：|2195099|+++a++b++++c++d++++++++++e
0++a++1++NaN++A+2018-05-17
1++b++b++2.0++B+2018-05-17
2++c++c++3.0++B+2018-05-17
3++d++2++4.0++A+2018-05-17|code-block|syntax|javascript|2195100|2195101|您可以检查调用dtypes的类型：|2195102|2195103|object+b+object+c+float64+d+category+e+datetime64ns+dtype:+object|2195104|您可以使用items()方法列出字符串列，如果dt为object|unordered-list-item|2195105|
，则可以按|2195106|df.dtypes.items，dt+in+df.dtypes.items()中的items()，dt进行筛选|blockquote|2195107|2195108|2195109|或您可以使用select_dtypes来显示仅包含以下字符串的数据帧：|2195110|2195111|df.select_dtypes(include=object)+a+b+0+a+1+1+b+b+2+c+c+3+d+2|2195112|entityMap^0|0|2|6|N|6|Y|6|0|0|0|0|0|0|0|0|7|6|0|0|0|5|7|0|0|15|7|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1J|8|@$D|1K|E|1L|F|G]|$D|1M|E|1N|F|G]|$D|1O|E|1P|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|1R|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1T|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1U|8|@]|9|@]|A|$]]|$1|Q|3|R|5|S|7|1V|8|@]|9|@]|A|$T|U]]|$1|V|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]|$1|W|3|X|5|K|7|1X|8|@$D|1Y|E|1Z|F|G]]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|20|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|21|8|@]|9|@]|A|$]]|$1|11|3|12|5|13|7|22|8|@$D|23|E|24|F|G]]|9|@]|A|$]]|$1|14|3|15|5|6|7|25|8|@]|9|@]|A|$]]|$1|16|3|17|5|18|7|26|8|@$D|27|E|28|F|G]]|9|@]|A|$]]|$1|19|3|-4|5|6|7|29|8|@]|9|@]|A|$]]|$1|1A|3|-4|5|6|7|2A|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|K|7|2B|8|@]|9|@]|A|$]]|$1|1D|3|-4|5|6|7|2C|8|@]|9|@]|A|$]]|$1|1E|3|1F|5|6|7|2D|8|@]|9|@]|A|$]]|$1|1G|3|-4|5|6|7|2E|8|@]|9|@]|A|$]]]|1H|$]]

Notice that the above answers will include DateTime, TimeStamp, Category and other datatypes.

Using <code>object</code> is more restrictive (although I am not sure if other <code>dtypes</code> would also of <code>object</code> dtype):

<ol>
<li>Create the dataframe:

<pre><code>df = pd.DataFrame({
 'a': ['a','b','c','d'], 
 'b': [1, 'b', 'c', 2], 
 'c': [np.nan, 2, 3, 4], 
 'd': ['A', 'B', 'B', 'A'], 
 'e': pd.to_datetime('today')})
df['d'] = df['d'].astype('category')
</code></pre></li>
</ol>

That will look like this:

<pre><code> a b c d e
0 a 1 NaN A 2018-05-17
1 b b 2.0 B 2018-05-17
2 c c 3.0 B 2018-05-17
3 d 2 4.0 A 2018-05-17
</code></pre>

<ol start="2">
<li>You can check the types calling <code>dtypes</code>:

<pre><code>df.dtypes

a object
b object
c float64
d category
e datetime64[ns]
dtype: object
</code></pre></li>
<li>You can list the strings columns using the <code>items()</code> method and filtering by <code>object</code>:

<pre><code>&gt; [ col for col, dt in df.dtypes.items() if dt == object]
['a', 'b']
</code></pre></li>
<li>Or you can use select_dtypes to display a dataframe with only the strings:

<pre><code>df.select_dtypes(include=[object])
 a b
0 a 1
1 b b
2 c c
3 d 2
</code></pre></li>
</ol>

blocks|key|2195136|text|我使用两步法:首先确定是否为dtype==object，如果是，则获取第一行数据，以查看该列的数据是否为字符串。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2195137|c+=+'my_column_name'
if+df[c].dtype+==+object+and+isinstance(df.iloc[0][c],+str):
++++#+do+something+|code-block|syntax|javascript|2195138|entityMap^0|E|D|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I use a 2-step approach: first to determine if <code>dtype==object</code>, and then if so, I got the first row of data to see if that column's data was a string or not. 

<pre><code>c = 'my_column_name'
if df[c].dtype == object and isinstance(df.iloc[0][c], str):
 # do something 
</code></pre>

blocks|key|5724427|text|这个问题已经提出4年了，我相信仍然没有一个明确的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5724428|我不认为字符串在熊猫(甚至是>=+1.0.0)中被视为一等公民。举个例子：|5724429|import+pandas+as+pd
import+datetime

df+=+pd.DataFrame({
++++'str':+['a',+'b',+'c',+None],
++++'hete':+[1,+2.0,+datetime.datetime.utcnow(),+None]
})

string_series+=+df['str']
print(string_series.dtype)
print(pd.api.types.is_string_dtype(string_series.dtype))

heterogenous_series+=+df['hete']
print(heterogenous_series.dtype)
print(pd.api.types.is_string_dtype(heterogenous_series.dtype))|code-block|syntax|javascript|5724430|打印|5724431|object
True
object
True|5724432|因此，尽管hete不包含任何显式字符串，但它被视为一个字符串序列。|offset|length|style|CODE|5724433|在阅读了the+documentation之后，我认为确保序列只包含字符串的唯一方法是：|5724434|def+is_string_series(s+:+pd.Series):
++++if+isinstance(s.dtype,+pd.StringDtype):
++++++++#+The+series+was+explicitly+created+as+a+string+series+(Pandas>=1.0.0)
++++++++return+True
++++elif+s.dtype+==+'object':
++++++++#+Object+series,+check+each+value
++++++++return+all((v+is+None)+or+isinstance(v,+str)+for+v+in+s)
++++else:
++++++++return+False|5724435|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/user_guide/text.html^0|0|0|0|0|0|5|4|0|4|H|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|18|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|19|8|@$O|1A|P|1B|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1C|8|@]|9|@$O|1D|P|1E|1|1F]]|A|$]]|$1|U|3|V|5|F|7|1G|8|@]|9|@]|A|$G|H]]|$1|W|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]]]

4 years since the creation of this question and I believe there's still not a definitive answer.
I don't think strings were ever considered as a first class citizen in Pandas (even &gt;= 1.0.0). As an example:
<pre><code>import pandas as pd
import datetime

df = pd.DataFrame({
 'str': ['a', 'b', 'c', None],
 'hete': [1, 2.0, datetime.datetime.utcnow(), None]
})

string_series = df['str']
print(string_series.dtype)
print(pd.api.types.is_string_dtype(string_series.dtype))

heterogenous_series = df['hete']
print(heterogenous_series.dtype)
print(pd.api.types.is_string_dtype(heterogenous_series.dtype))
</code></pre>
prints
<pre><code>object
True
object
True
</code></pre>
so although <code>hete</code> does not contain any explicit strings, it is considered as a string series.
After reading <a href="https://pandas.pydata.org/pandas-docs/stable/user_guide/text.html" rel="nofollow noreferrer">the documentation</a>, I think the only way to make sure a series contains only strings is:
<pre><code>def is_string_series(s : pd.Series):
 if isinstance(s.dtype, pd.StringDtype):
 # The series was explicitly created as a string series (Pandas&gt;=1.0.0)
 return True
 elif s.dtype == 'object':
 # Object series, check each value
 return all((v is None) or isinstance(v, str) for v in s)
 else:
 return False
</code></pre>

blocks|key|1717220|text|据我所知，唯一确定了解类型的方法是检查值，然后你可以做一个断言，看看它是不是你想要的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1717221|下面的函数将获取列中每个值的数据类型，删除重复项，然后转换为一个列表，以便您可以查看它/与之交互。这将使您能够以您希望的方式处理混合类型、对象和NAs+(当然，np.nan是浮点类型，但我将这些内容留给感兴趣的读者)|1717222|import+pandas+as+pd

df+=+pd.DataFrame({"col1":+[1,+2,+3,+4],
+++++++++++++++++++"col2":+["a",+"b",+"c",+"d"],
+++++++++++++++++++"col3":+[[1,+2,+3,+4],+[1,+2,+3,+4],+[1,+2,+3,+4],+[1,+2,+3,+4]]
+++++++++++++++++++})

print(df.dtypes.to_dict())
#+{'col1':+dtype('int64'),+'col2':+dtype('O'),+'col3':+dtype('O')}

def+true_dtype(df):+#+You+could+add+a+column+filter+here+too
++++return+{col:+df[col].apply(lambda+x:+type(x)).unique().tolist()+for+col+in+df.columns}

true_types+=+true_dtype(df)
print(true_types)
#+{'col1':+[<class+'int'>],+'col2':+[<class+'str'>],+'col3':+[<class+'list'>]}

print(true_types['col2']+==+[str])
#+True|code-block|syntax|javascript|1717223|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

As far as I can tell, the only sure fire way to know what types are there is to check the values, then you can do an assertion to see if it's what you expect.
The below function will get the dtypes of each value in a column, drop duplicates and then cast to a list so you can view/interact with it. This will let you deal with mixed types, objects and NAs the way you wish (of course np.nan is of type float but I leave such things to the interested reader)
<pre><code>import pandas as pd

df = pd.DataFrame({&quot;col1&quot;: [1, 2, 3, 4],
 &quot;col2&quot;: [&quot;a&quot;, &quot;b&quot;, &quot;c&quot;, &quot;d&quot;],
 &quot;col3&quot;: [[1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4]]
 })

print(df.dtypes.to_dict())
# {'col1': dtype('int64'), 'col2': dtype('O'), 'col3': dtype('O')}

def true_dtype(df): # You could add a column filter here too
 return {col: df[col].apply(lambda x: type(x)).unique().tolist() for col in df.columns}

true_types = true_dtype(df)
print(true_types)
# {'col1': [&lt;class 'int'&gt;], 'col2': [&lt;class 'str'&gt;], 'col3': [&lt;class 'list'&gt;]}

print(true_types['col2'] == [str])
# True
</code></pre>

blocks|key|5724470|text|这将返回dtype为string的列名的列表(本例中为object)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5724471|#let+df+be+your+dataframe+++++
df.columns[df.dtypes==object].tolist()|code-block|syntax|javascript|5724472|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This will return a list of column name whose dtype is string(object in this case)
<pre><code>#let df be your dataframe 
df.columns[df.dtypes==object].tolist()
</code></pre>

blocks|key|2195190|text|在Pandas1.0中，引入了convert_dtypes。当列没有显式创建为StringDtype时，可以很容易地对其进行转换。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2195191|然后，pd.StringDtype.is_dtype将为wtring列返回True。即使它们包含NA值也是如此。|2195192|对于旧的和新的样式字符串，完整的检查序列可能如下所示：|2195193|def+has_string_type(s:+pd.Series)+->+bool:
++++if+pd.StringDtype.is_dtype(s.dtype):
++++++++#+StringDtype+extension+type
++++++++return+True

++++if+s.dtype+!=+"object":
++++++++#+No+object+column+-+definitely+no+string
++++++++return+False

++++try:
++++++++s.str
++++except+AttributeError:
++++++++return+False

++++#+The+str+accessor+exists,+this+must+be+a+String+column
++++return+False|code-block|syntax|javascript|2195194|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.convert_dtypes.html|1|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.StringDtype.html#pandas.StringDtype^0|F|E|13|B|F|E|0|13|B|1|0|3|N|11|4|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]|$9|16|A|17|1|18]]|E|$]]|$1|F|3|G|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1E|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1F|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

With Pandas 1.0 <a href="https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.convert_dtypes.html" rel="nofollow noreferrer"><code>convert_dtypes</code></a> was introduced. When a column was not explicitly created as <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.StringDtype.html#pandas.StringDtype" rel="nofollow noreferrer"><code>StringDtype</code></a> it can be easily converted.
<code>pd.StringDtype.is_dtype</code> will then return <code>True</code> for wtring columns. Even when they contain NA values.
For old and new style strings the complete series of checks could be something like this:
<pre><code>def has_string_type(s: pd.Series) -&gt; bool:
 if pd.StringDtype.is_dtype(s.dtype):
 # StringDtype extension type
 return True

 if s.dtype != &quot;object&quot;:
 # No object column - definitely no string
 return False

 try:
 s.str
 except AttributeError:
 return False

 # The str accessor exists, this must be a String column
 return False
</code></pre>

I want check if columns in a dataframe consists of strings so I can label them with numbers for machine learning purposes. Some columns consists of numbers, I dont want to change them. Columns example can be seen below:

<pre><code>TRAIN FEATURES
 Age Level 
 32.0 Silver 
 61.0 Silver 
 66.0 Silver 
 36.0 Gold 
 20.0 Silver 
 29.0 Silver 
 46.0 Silver 
 27.0 Silver 
</code></pre>

Thank you=)

Python: Check if dataframe column contain string type

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

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我想检查dataframe中的列是否由字符串组成，这样我就可以用数字来标记它们，以便进行机器学习。有些列是由数字组成的，我不想更改它们。列示例如下：TRAIN FEATURES  Age              Level    32.0              Silver        61.0        ...

问Python:检查dataframe列是否包含字符串类型
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问Python:检查dataframe列是否包含字符串类型EN