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Python - Create list with numbers between 2 values?
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Use range. In Python 2.x it returns a list so all you need is:
>>> range(11, 17)
[11, 12, 13, 14, 15, 16]
In Python 3.x range is a iterator. So, you need to convert it to a list:
>>> list(range(11, 17))
[11, 12, 13, 14, 15, ...
Reverse Range in Swift
...sed() method on a range
for i in (1...5).reversed() { print(i) } // 5 4 3 2 1
Or stride(from:through:by:) method
for i in stride(from:5,through:1,by:-1) { print(i) } // 5 4 3 2 1
stide(from:to:by:) is similar but excludes the last value
for i in stride(from:5,to:0,by:-1) { print(i) } // 5 4 3...
oracle10g 网址收藏 - ORACLE - 清泛IT论坛，有思想、有深度
...迅雷进行下载，就不用登陆OTN了：
Oracle Database 10g Release 2 (10.2.0.1.0) Enterprise/Standard Edition for Microsoft Windows (32-bit)
http://download.oracle.com/otn/nt/oracle10g/10201/10201_database_win32.zip
http://download.oracle.com/otn/nt/oracle10g/10201/10201_client_win32.zip
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轻松学习App开发？App Inventor 2 中文网搞定！ - App Inventor 2 中文网 -...
轻松学习App开发？App Inventor 2 中文网搞定！chatgpt_ai2有没有想过自己动手开发一个属于自己的应用程序？有没有因为开发难度而望而却步？那么现在，我有一个好消息要告诉你，App Inventor 2 中文网（fun123 cn）能帮你搞定！App Inv ...
Could not calculate build plan: Plugin org.apache.maven.plugins:maven-resources-plugin:2.5 or one of
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29 Answers
29
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How to calculate the number of occurrence of a given character in each row of a column of strings?
...a")
q.data
# number string number.of.a
#1 1 greatgreat 2
#2 2 magic 1
#3 3 not 0
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How can I check if a program exists from a Bash script?
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How to insert element into arrays at specific position?
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23 Answers
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numpy: most efficient frequency counts for unique values in an array
...ence/generated/numpy.bincount.html
import numpy as np
x = np.array([1,1,1,2,2,2,5,25,1,1])
y = np.bincount(x)
ii = np.nonzero(y)[0]
And then:
zip(ii,y[ii])
# [(1, 5), (2, 3), (5, 1), (25, 1)]
or:
np.vstack((ii,y[ii])).T
# array([[ 1, 5],
[ 2, 3],
[ 5, 1],
[25, ...
dropping infinite values from dataframes in pandas?
... dropna:
df.replace([np.inf, -np.inf], np.nan).dropna(subset=["col1", "col2"], how="all")
For example:
In [11]: df = pd.DataFrame([1, 2, np.inf, -np.inf])
In [12]: df.replace([np.inf, -np.inf], np.nan)
Out[12]:
0
0 1
1 2
2 NaN
3 NaN
The same method would work for a Series.
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