blocks|key|1459663|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1459664|df.applymap(lambda+x:+float(x.replace(",",+".")))|code-block|syntax|javascript|1459665|编辑:忘记了@Psidom显示的map|offset|length|style|CODE|1459666|entityMap^0|0|0|G|3|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]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Try this:

<pre><code>df.applymap(lambda x: float(x.replace(",", ".")))
</code></pre>

Edit: forgot the <code>map</code> as @Psidom showed

blocks|key|393879|text|您可以使用应用程序图|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|393880|df[["Lon_X",+"Lat_Y"]]+=+df[["Lon_X",+"Lat_Y"]].applymap(lambda+x:+float(x.replace(",",+".")))
df|code-block|syntax|javascript|393881|​|393882|📷|atomic|393883|393884|下面是关于这些替代方法的一些基准，to_float_inplace比所有其他方法都要快得多：|style|CODE|393885|数据|393886|df+=+pd.DataFrame({"Lon_X":+["5,234234"+for+i+in+range(1000000)],+"Lat_Y":+["6,3234234"+for+i+in+range(1000000)]})|393887|#+to_float_inplace
def+to_float_inplace(x):
++++x[:]+=+x.str.replace(',',+'.').astype(float)

%25timeit+df.apply(to_float_inplace)
#+1+loops,+best+of+3:+269+ms+per+loop

#+applymap+%2B+astype
%25timeit+df.applymap(lambda+x:+x.replace(",",+".")).astype(float)
#+1+loops,+best+of+3:+1.26+s+per+loop

#+to_float
def+to_float(x):
++++return+x.str.replace(',',+'.').astype(float)

%25timeit+df.apply(to_float)
#+1+loops,+best+of+3:+1.47+s+per+loop

#+applymap+%2B+float
%25timeit+df.applymap(lambda+x:+float(x.replace(",",+".")))
#+1+loops,+best+of+3:+1.75+s+per+loop

#+replace+with+regex
%25timeit+df.replace(',',+'.',+regex=True).astype(float)
#+1+loops,+best+of+3:+1.79+s+per+loop|393888|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.applymap.html|1|IMAGE|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/dd202b87a6b38970274d8f5d221a5951.jpg|imageAlt^0|5|5|0|0|0|0|0|1|1|0|0|H|G|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@$A|1D|B|1E|1|1F]]|C|$]]|$1|D|3|E|5|F|7|1G|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1H|8|@]|9|@]|C|$]]|$1|K|3|L|5|M|7|1I|8|@]|9|@$A|1J|B|1K|1|1L]]|C|$]]|$1|N|3|J|5|6|7|1M|8|@]|9|@]|C|$]]|$1|O|3|P|5|6|7|1N|8|@$A|1O|B|1P|Q|R]]|9|@]|C|$]]|$1|S|3|T|5|6|7|1Q|8|@]|9|@]|C|$]]|$1|U|3|V|5|F|7|1R|8|@]|9|@]|C|$G|H]]|$1|W|3|X|5|F|7|1S|8|@]|9|@]|C|$G|H]]|$1|Y|3|-4|5|6|7|1T|8|@]|9|@]|C|$]]]|Z|$10|$5|11|12|13|C|$14|15]]|16|$5|17|12|18|C|$19|1A|1B|-4]]]]

You can use <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.applymap.html" rel="nofollow noreferrer">applymap</a>:

<pre><code>df[["Lon_X", "Lat_Y"]] = df[["Lon_X", "Lat_Y"]].applymap(lambda x: float(x.replace(",", ".")))
df
</code></pre>

<a href="https://i.stack.imgur.com/vJpgL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vJpgL.jpg" alt="enter image description here"></a>

<hr>

Here is some benchmark about these alternatives, <code>to_float_inplace</code> is significantly faster than all other methods:

Data:

<pre><code>df = pd.DataFrame({"Lon_X": ["5,234234" for i in range(1000000)], "Lat_Y": ["6,3234234" for i in range(1000000)]})
</code></pre>

<hr>

<pre><code># to_float_inplace
def to_float_inplace(x):
 x[:] = x.str.replace(',', '.').astype(float)

%timeit df.apply(to_float_inplace)
# 1 loops, best of 3: 269 ms per loop

# applymap + astype
%timeit df.applymap(lambda x: x.replace(",", ".")).astype(float)
# 1 loops, best of 3: 1.26 s per loop

# to_float
def to_float(x):
 return x.str.replace(',', '.').astype(float)

%timeit df.apply(to_float)
# 1 loops, best of 3: 1.47 s per loop

# applymap + float
%timeit df.applymap(lambda x: float(x.replace(",", ".")))
# 1 loops, best of 3: 1.75 s per loop

# replace with regex
%timeit df.replace(',', '.', regex=True).astype(float)
# 1 loops, best of 3: 1.79 s per loop
</code></pre>

blocks|key|1261424|text|您可以将熊猫的矢量化方法应用到一个轴上：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1261425|def+to_float_inplace(x):
++++x[:]+=+x.str.replace(',',+'.').astype(float)

df.apply(to_float_inplace)|code-block|syntax|javascript|1261426|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|$]]

You can apply panda's vectorized methods along an axis in-place:

<pre><code>def to_float_inplace(x):
 x[:] = x.str.replace(',', '.').astype(float)

df.apply(to_float_inplace)
</code></pre>

blocks|key|1459764|text|可以跳过使用apply，直接用replace方法替换为regex=True|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1459765|df.replace(',',+'.',+regex=True).astype(float)|code-block|syntax|javascript|1459766|entityMap^0|F|7|R|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can skip using apply and replace directly with the <code>replace</code> method with <code>regex=True</code>

<pre><code>df.replace(',', '.', regex=True).astype(float)
</code></pre>

blocks|key|1261444|text|令人惊讶的是，在np系列上迭代比使用pd.series.str.replace更快。我用2m行系列做了下面的实验|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1261445|setup+=+'''
import+pandas+as+pd
import+numpy+as+np
a+=+pd.Series(list('aabc')+*+500000)
b+=+a.values.astype(str)
'''

a+=+'''
a[:]+=+a.str.replace("b",+"d")
'''
b+=+'''
b[:]+=+np.char.replace(b,+"b",+"d")
'''
c+=+'''
for+i,+x+in+enumerate(b):
++++if+"b"+in+x:
++++++++b[i]+=+"d"
'''
a_speed+=+min(timeit.Timer(a,+setup=setup).repeat(7,+5))
b_speed+=+min(timeit.Timer(b,+setup=setup).repeat(7,+5))
c_speed+=+min(timeit.Timer(c,+setup=setup).repeat(7,+5))|code-block|syntax|javascript|1261446|结果：|1261447|a_speed+=+2.3304627019997497|1261448|b_speed+=+6.832672896000076|1261449|c_speed+=+1.9407824309996613|1261450|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Surprisingly, it appears to be faster to iterate over an np series rather than use pd.series.str.replace. I did the following experiment with a 2m row series

<pre><code>setup = '''
import pandas as pd
import numpy as np
a = pd.Series(list('aabc') * 500000)
b = a.values.astype(str)
'''

a = '''
a[:] = a.str.replace("b", "d")
'''
b = '''
b[:] = np.char.replace(b, "b", "d")
'''
c = '''
for i, x in enumerate(b):
 if "b" in x:
 b[i] = "d"
'''
a_speed = min(timeit.Timer(a, setup=setup).repeat(7, 5))
b_speed = min(timeit.Timer(b, setup=setup).repeat(7, 5))
c_speed = min(timeit.Timer(c, setup=setup).repeat(7, 5))
</code></pre>

result:

a_speed = 2.3304627019997497

b_speed = 6.832672896000076

c_speed = 1.9407824309996613

<pre><code>Lon_X Lat_Y
5,234234 6,3234234
5,234234 6,3234234
5,234234 6,3234234
5,234234 6,3234234
5,234234 6,3234234
</code></pre>

I've GPS coordinates in a pandas/dataframe like above. These however use the comma separator. What's the best way using pandas to convert these to float GPS coordinates?

<pre><code>for item in frame.Lon_X:
 float(item.replace(",", ".")) # makes the conversion but does not store it back
</code></pre>

I've tried the iteritems function, but seems very slow and gives me a warning that I don't really understand:

<pre><code>for index, value in frame.Lon_X.iteritems():
 frame.Lon_X[index] = float(value.replace(",", "."))
</code></pre>

<blockquote>
 See the caveats in the documentation:
 <a href="http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy" rel="nofollow noreferrer">http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy</a>
 from ipykernel import kernelapp as app
</blockquote>

Replace values in dataframe Python

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。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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Lon_X        Lat_Y5,234234     6,32342345,234234     6,32342345,234234     6,32342345,234234     6,32342345,234234     6,3234234我有像上面那样的熊猫/数据中心的GPS坐标。但是，它们使用逗号分...

问替换dataframe Python中的值
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问替换dataframe Python中的值EN